Suppose $m\geq 3$. In the case that $m$ is odd for $U = C^nZ_m$, from the proof of Theorem~\ref{thm:cnzm}, we obtain that $\phi(\mathsf{W}_X)$ is spanned by $\{e_i, 0\leq i\leq 2^{n+1}-1\}$. It means that $\phi(\mathsf{W}_X)$ is equal to $\mathbb{Z}_{m}^{2^{n+1}}$. Note that $\phi(C^nZ_m) = e_{2^{n+1}-1}$, so in this case the target gate $U$ itself belongs to $\mathsf{W}_X$ and hence $U$ belongs to the twirling group $\mathsf{G} = \mathsf{X}\ltimes \mathsf{W}_X$. When $U\in \mathsf{G}$, then
\begin{equation}
\begin{split}
\mathcal{U}\Lambda_G \mathcal{U}^{\dagger} &= \mathbb{E}_{G\in \mathsf{G}} \mathcal{U}\mathcal{G}\Lambda \mathcal{G}^{\dagger}\mathcal{U}^{\dagger}\\
&= \mathbb{E}_{G\in \mathsf{G}} \mathcal{G}\Lambda \mathcal{G}^{\dagger}\\
&= \Lambda_G.
\end{split}
\end{equation}
The commutation between $\mathcal{U}$ and $\Lambda_G$ is satisfied automatically.

Below we discuss the case that $m$ is even and $m\geq 3$ for $U=C^nZ_m$. Based on Lemma~\ref{lemma:czdihedral}, we study whether $CZ$ and $S$ gates belong to $\mathsf{W}_X$ or not. By symmetry, we only need to study that for the phase gate on the first qubit $S_1$ and the controlled-phase gate on the first two qubits $CZ_{12}$. Note that the phase gate $S$ can be expressed as $S=Z_m^{m/4}$ and $CZ$ can be expressed as $CZ=CZ_{m}^{m/2}$. We express $m=q2^k$ where $q\geq 1$ is odd and $k\geq 1$ is a positive integer. Note that the bases of $\phi(\mathsf{W}_X)$ are $\{v_i=e_i-e_{2^{n+1}-1}, 0\leq i\leq 2^{n+1}-2\}$ along with $u = 2^{\min(n+1,k)}e_{2^{n+1}-1}$. Then, $\phi(S_1)$ can be constructed by
\begin{equation}\label{eq:S1}
\begin{split}
\phi(S_1) &= q2^{k-2}(e_{2^{n+1}-2^n} + e_{2^{n+1}-2^n+1} + \cdots + e_{2^{n+1}-1})\\
&=q2^{k-2}(v_{2^{n+1}-2^n} + v_{2^{n+1}-2^n+1} + \cdots + v_{2^{n+1}-2}) + q2^{k-2}(2^n-1)e_{2^{n+1}-1} + q2^{k-2}e_{2^{n+1}-1}\\
&=q2^{k-2}(v_{2^{n+1}-2^n} + v_{2^{n+1}-2^n+1} + \cdots + v_{2^{n+1}-2}) + q2^{n+k-2}e_{2^{n+1}-1}\\
&= q2^{k-2}(v_{2^{n+1}-2^n} + v_{2^{n+1}-2^n+1} + \cdots + v_{2^{n+1}-2}) + q2^{n+k-2-\min(n+1, k)}u\\
&= q2^{k-2}(v_{2^{n+1}-2^n} + v_{2^{n+1}-2^n+1} + \cdots + v_{2^{n+1}-2}) + q2^{\max(k-3, n-2)}u.
\end{split}
\end{equation}
Meanwhile, $\phi(CZ_{12})$ can be constructed by
\begin{equation}\label{eq:CZ12}
\begin{split}
\phi(CZ_{12}) &= q2^{k-1}(e_{2^{n+1}-2^{n-1}} + e_{2^{n+1}-2^{n-1}+1} + \cdots + e_{2^{n+1}-1})\\
&=q2^{k-1}(v_{2^{n+1}-2^{n-1}} + v_{2^{n+1}-2^{n-1}+1} + \cdots + v_{2^{n+1}-2}) + q2^{k-1}(2^{n-1}-1)e_{2^{n+1}-1} + q2^{k-1}e_{2^{n+1}-1}\\
&=q2^{k-1}(v_{2^{n+1}-2^{n-1}} + v_{2^{n+1}-2^{n-1}+1} + \cdots + v_{2^{n+1}-2}) + q2^{n+k-2}e_{2^{n+1}-1}\\
&= q2^{k-1}(v_{2^{n+1}-2^{n-1}} + v_{2^{n+1}-2^{n-1}+1} + \cdots + v_{2^{n+1}-2}) + q2^{n+k-2-\min(n+1, k)}u\\
&= q2^{k-1}(v_{2^{n+1}-2^{n-1}} + v_{2^{n+1}-2^{n-1}+1} + \cdots + v_{2^{n+1}-2}) + q2^{\max(k-3, n-2)}u.
\end{split}
\end{equation}
The constructions Eq.~\eqref{eq:S1} and Eq.~\eqref{eq:CZ12} are valid only when the coefficients of the bases are integers. Obviously, $k\geq 3$, and $k = 2$ and $n\geq 2$ can make the coefficients to be integers. In this case, $S_1$ and $CZ_{12}$ belong to $\mathsf{W}_X$ and furthermore, $\langle CZ, X, S\rangle \leq \mathsf{G}$. Based on Lemma~\ref{lemma:czdihedral}, $\mathcal{U}\Lambda_G \mathcal{U}^{\dagger} = \Lambda_G$ is satisfied. The exceptions are the case that $k = 1$ and that $n = 1$ and $k = 2$. Below we discuss the two kinds of exceptions in detail.

When $k = 1$, the target gate $U = C^nZ_m$ and $m=2q$. In this case, $u=2e_{2^{n+1}-1}$, and $CZ_{m/2}$ and $Z_m$ gates belong to $\mathsf{G}$, which can be derived through the following equations.
\begin{equation}
\begin{split}
\phi((Z_m)_1) &= e_{2^{n+1}-2^n} + e_{2^{n+1}-2^n+1} + \cdots + e_{2^{n+1}-1}\\
&= v_{2^{n+1}-2^n} + v_{2^{n+1}-2^n+1} + \cdots + v_{2^{n+1}-2} + (2^n-1)e_{2^{n+1}-1} + e_{2^{n+1}-1}\\
&= v_{2^{n+1}-2^n} + v_{2^{n+1}-2^n+1} + \cdots + v_{2^{n+1}-2} + 2^ne_{2^{n+1}-1}\\
&= v_{2^{n+1}-2^n} + v_{2^{n+1}-2^n+1} + \cdots + v_{2^{n+1}-2} + 2^{n-1}u.
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\phi((CZ_{m/2})_{12}) &= 2(e_{2^{n+1}-2^{n-1}} + e_{2^{n+1}-2^{n-1}+1} + \cdots + e_{2^{n+1}-1})\\
&=2(v_{2^{n+1}-2^{n-1}} + v_{2^{n+1}-2^{n-1}+1} + \cdots + v_{2^{n+1}-2}) + 2(2^{n-1}-1)e_{2^{n+1}-1} + 2e_{2^{n+1}-1}\\
&=2(v_{2^{n+1}-2^{n-1}} + v_{2^{n+1}-2^{n-1}+1} + \cdots + v_{2^{n+1}-2}) + 2^ne_{2^{n+1}-1}\\
&= 2(v_{2^{n+1}-2^{n-1}} + v_{2^{n+1}-2^{n-1}+1} + \cdots + v_{2^{n+1}-2}) + 2^{n-1}u.
\end{split}
\end{equation}
Thus, when $k=1$, $\langle CZ_{m/2}, Z_m \rangle \leq \mathsf{W}_X$ and $\mathsf{X}\ltimes \langle CZ_{m/2}, Z_m \rangle \leq \mathsf{G}$. Notice that $m\geq 3$ and $m/2$ is odd, so $m/2\geq 3$ holds. It can be verified that the irreducible representation decomposition of the Liouville representation for group $\langle CZ_m, Z_m, X \rangle$, where $m\geq 3$ and $m$ is odd, and group $\langle CZ, S, X \rangle$ are the same. As $\mathsf{X}\ltimes \langle CZ_{m/2}, Z_{m/2} \rangle \leq \mathsf{X}\ltimes \langle CZ_{m/2}, Z_m \rangle \leq \mathsf{G}$ and $m/2\geq 3$, after being twirled by $\mathsf{G}$, the twirled noise channel $\Lambda_G$ would have the same blocks as $\mathcal{U}$, and $\mathcal{U}\Lambda_G \mathcal{U}^{\dagger} = \Lambda_G$ is satisfied.

When $n = 1$ and $k = 2$, the target gate $U = CZ_m$ and $m=4q$. The bases for $\phi(\mathsf{W}_X)$ are $v_0 = (1, 0, 0, -1)^T$, $v_1 = (0, 1, 0, -1)^T$, $v_2 = (0, 0, 1, -1)^T$, and $u = (0, 0, 0, 4)$. Note that $\phi(CZ\cdot S_1) = (0, 0, q, 3q)^T = v_2+u$ and $\phi(CZ\cdot S_2) = (0, q, 0, 3q)^T = v_1+u$. Thus, in this case, $\langle CZ\cdot S_1, CZ\cdot S_2 \rangle \in \mathsf{W}_X \leq \mathsf{G}$. In the proof of Theorem~\ref{thm:cnzm}, we have already shown that Pauli group $\mathsf{P}_2\in \mathsf{G}$. Thus, the twirled noise channel would be diagonal in Liouville representation, $\Lambda_G = \sum_i \lambda_i \lketbra{\sigma_i}{\sigma_i}$. Also, we have the following identities,
\begin{align}
CZ\cdot S_1 X_1 (CZ\cdot S_1)^{\dagger} &= X_1 Z_1 Z_2;\\
CZ\cdot S_2 X_1 (CZ\cdot S_2)^{\dagger} &= X_1 Z_2;\\
(CZ\cdot S_2 \cdot CZ\cdot S_1) X_1 (CZ\cdot S_2 \cdot CZ\cdot S_1)^{\dagger} &= X_1 Z_1.
\end{align}
Under the twirling of $\langle CZ\cdot S_1, CZ\cdot S_2 \rangle$, $X_1$, $X_1 Z_1$, $X_1 Z_2$, and $X_1 Z_1Z_2$ would be symmetric. Their corresponding Pauli fidelities $\lambda_i$ would be the same after the twirling. As the twirling group contains $\langle CZ\cdot S_1, CZ\cdot S_2 \rangle$, the twirled noise channel $\Lambda_G$ would be diagonal and proportional to identity in the space spanned by $X_1\mathsf{Z}$. The cases for spaces spanned by $X_2\mathsf{Z}$ or $X_1X_2\mathsf{Z}$ are the same. In summary, $\Lambda_G$ has diagonal blocks in spaces spanned by $X_1\mathsf{Z}$, $X_2\mathsf{Z}$ and $X_1X_2\mathsf{Z}$, respectively. Thus, in this case, $\mathcal{U}\Lambda_G \mathcal{U}^{\dagger} = \Lambda_G$ is also satisfied.

At last, we analyze the case that $m=2$ and $n\geq 2$. The case that $m=2$ and $n=1$ is just CZ and obviously in this case $\mathcal{U}\Lambda_G \mathcal{U}^{\dagger} \neq \Lambda_G$. When the target gate is $C^{n}Z$ with $n\geq 2$, the twirling group is $\mathsf{G}=X\ltimes W_X=\langle C^{n-1}Z,\cdots, Z, X\rangle$. Different from a generic diagonal gate, the diagonal blocks spanned by $X_{\mathbf{a}}\mathsf{Z}$ of $C^{n}Z$ gate in Pauli-Liouville representation can be further divided into two small blocks. We will show that any channel twirled by group $\langle CZ, Z, X\rangle$ would have the same diagonal blocks as $C^{n}Z$ in Liouville representation. As $\langle C^{n-1}Z,\cdots, Z, X\rangle$ contains $\langle CZ, Z, X\rangle$, $\Lambda_{\langle C^{n-1}Z,\cdots, Z, X\rangle}$ would commute with $C^nZ$.

Similar to Eq.~\eqref{eq:PauliOfU}, by setting $\sigma_i = \frac{1}{\sqrt{d}} X_{\mathbf{a}_i}Z_{\mathbf{b}_i}$ and $\sigma_j = \frac{1}{\sqrt{d}} X_{\mathbf{a}_j}Z_{\mathbf{b}_j}$ where $X_{\mathbf{a}_i}, X_{\mathbf{a}_j}\in \{\mathbb{I}, X\}^{\otimes n}$ and $Z_{\mathbf{b}_i}, Z_{\mathbf{b}_j}\in \{\mathbb{I}, Z\}^{\otimes n}$, we get the matrix element of the Liouville representation of $U=C^nZ$,
\begin{equation}
\begin{split}
\mathcal{U}_{ji}&=d\tr(\sigma_i\sigma_jC^{n}Z\sigma_iC^{n}Z\sigma_i^{\dagger})\\
&\propto \frac{1}{d}\tr(X_{\mathbf{a}_i}X_{\mathbf{a}_j}Z_{\mathbf{b}_i}Z_{\mathbf{b}_j}W);\\
W&=C^{n}Z X_{\mathbf{a}_i} C^{n}Z X_{\mathbf{a}_i}.
\end{split}
\end{equation}
Notice that to make $\mathcal{U}_{ji}$ nonzero, we at least require $X_{\mathbf{a}_i}=X_{\mathbf{a}_j}$. Define support of $X_{\mathbf{a}_i}$ to be the set, $\text{supp}(X_{\mathbf{a}_i})=\{i|X_{\mathbf{a}_i}\text{ on qubit $i$ is }X\}$, which is essentially the location of $1$ in $\mathbf{a}_i$. Notice that two of the diagonal elements of $W$ are $-1$ while others are all $1$. Suppose these two elements are $W_{i_1,i_1}$ and $W_{i_2,i_2}$. Without loss of generality, $i_2$ can be set as $2^{n+1}-1$ as the element of the last row and last column is $-1$. To make $\mathcal{U}_{ji}$ nonzero, we furthermore require that $(Z_{\mathbf{b}_i}Z_{\mathbf{b}_j})_{i_1,i_1}=(Z_{\mathbf{b}_i}Z_{\mathbf{b}_j})_{i_2,i_2}$, or $\bra{i_1}Z_{\mathbf{b}_i}Z_{\mathbf{b}_j}\ket{i_1}=\bra{i_2}Z_{\mathbf{b}_i}Z_{\mathbf{b}_j}\ket{i_2}$. Note that $\ket{i_2} = X_{\mathbf{a}_i}\ket{i_1}$. Thus, to make $\mathcal{U}_{ji}$ nonzero, we require that $X_{\mathbf{a}_i}$ and $Z_{\mathbf{b}_i}Z_{\mathbf{b}_j}$ commute, which is equivalent to that $Z_{\mathbf{b}_i}Z_{\mathbf{b}_j}$ has even number of $Z$ gates on qubits of $\text{supp}(X_{\mathbf{a}_i})$. Hence, the block spanned by $X_{\mathbf{a}_i}\mathsf{Z}$ can be further split into two blocks when $m=2$. One block is spanned by $X_{\mathbf{a}_i}\mathsf{Z}_e = \{X_{\mathbf{a}_i}Z_{\mathbf{b}_k} , \mod(\abs{\mathbf{a}_i\cap \mathbf{b}_k}, 2)=0 \}$ and the other block is spanned by $X_{\mathbf{a}_i}\mathsf{Z}_o = \{X_{\mathbf{a}_i}Z_{\mathbf{b}_k} , \mod(\abs{\mathbf{a}_i\cap \mathbf{b}_k}, 2)=1\}$. Here, $\mathbf{a}_i\cap \mathbf{b}_k$ is the bitwise and operation between $\mathbf{a}_i$ and $\mathbf{b}_k$, $\abs{\mathbf{a}_i\cap \mathbf{b}_k}$ is the weight of $\mathbf{a}_i\cap \mathbf{b}_k$, and $\mod(\abs{\mathbf{a}_i\cap \mathbf{b}_k}, 2)=0$ means that there are even number of $Z$ gates of $Z_{\mathbf{b}_k}$ acting on qubits of $\text{supp}(X_{\mathbf{a}_i})$. Notice that $\mod(\abs{\mathbf{a}_i\cap \mathbf{b}_k}, 2)=0$ is equivalent to that $\sum_{1\leq j\leq n} (\mathbf{a}_i)_j (\mathbf{b}_k)_j$ is even and $\mod(\abs{\mathbf{a}_i\cap \mathbf{b}_k}, 2)=1$ is equivalent to that $\sum_{1\leq j\leq n} (\mathbf{a}_i)_j (\mathbf{b}_k)_j$ is odd. These relations are useful in the following proof.

Now we obtain that the Liouville representation of $\mathcal{U}$ is spanned by blocks with bases $X_{\mathbf{a}}\mathsf{Z}_e$ and $X_{\mathbf{a}}\mathsf{Z}_o$. To show $\mathcal{U}\Lambda_G \mathcal{U}^{\dagger} = \Lambda_G$, we also need to check whether $\Lambda_G$ has the same diagonal blocks with $\mathcal{U}$ and whether $\Lambda_G$ is proportional to identity when restricting in these blocks. The proof is similar to that in Lemma~\ref{lemma:czdihedral}. Recall that $\Lambda_G = \sum_i \lambda_i \lketbra{\sigma_i}{\sigma_i}$ in Liouville representation. Also, if there exists an element $G\in \mathsf{G}$ such that $\sigma_i=G\sigma_jG^{\dagger}$, then $\lambda_i=\lambda_j$, and $\sigma_i$ and $\sigma_j$ would be in the same blocks. Then, we only need to check whether any element in $X_{\mathbf{a}}\mathsf{Z}_e$ can be generated from $X_{\mathbf{a}}$ by conjugate action of elements in $\mathsf{G}$.

Given $X_{\mathbf{a}}\in \{\mathbb{I},X\}^{\otimes n} / \mathbb{I}^{\otimes n}$, choose $i$ from $\text{supp}(X_{\mathbf{a}})$. For any $Z_{\mathbf{b}}\in \{\mathbb{I},Z\}^{\otimes n}$ where $X_{\mathbf{a}}Z_{\mathbf{b}}\in X_{\mathbf{a}}\mathsf{Z}_e$, we can choose $W=\prod_{j\neq i}CZ_{ij}^{b_j}$, such that
\begin{equation}
\begin{split}
=&X_{\mathbf{a}}(\prod_{1\leq j\leq n, j\neq i}Z_j^{b_j}) Z_i^{\sum_{1\leq j\leq n, j\neq i} a_jb_j} \\
=&X_{\mathbf{a}}(\prod_{1\leq j\leq n, j\neq i}Z_j^{b_j}) Z_i^{(\sum_{1\leq j\leq n} a_jb_j)-a_ib_i}\\
=&X_{\mathbf{a}}(\prod_{1\leq j\leq n, j\neq i}Z_j^{b_j}) Z_i^{-a_ib_i}\\
=&X_{\mathbf{a}}(\prod_{1\leq j\leq n, j\neq i}Z_j^{b_j}) Z_i^{b_i}\\
\end{split}
\end{equation}
Here, in the fourth line, we utilize $X_{\mathbf{a}}Z_{\mathbf{b}}\in X_{\mathbf{a}}\mathsf{Z}_e$, which means $\sum_{1\leq j\leq n} a_jb_j$ is even. In the fifth line, we use the condition that $a_i=1$. Thus, we show that $\Lambda_G$ is proportional to identity when restricting in the block spanned by $X_{\mathbf{a}}\mathsf{Z}_e$. From a similar argument, it can be verified that this is also true for block spanned by $X_{\mathbf{a}}\mathsf{Z}_o$. Then, we successfully show that $\mathcal{U}$ and $\Lambda_G$ have the same diagonal blocks, and $\mathcal{U}\Lambda_G\mathcal{U}^{-1}=\Lambda_G$ when $m=2$.

As a conclusion, the twirling group $X\ltimes \mathsf{W}_X$ satisfies $\mathcal{U}\Lambda_G\mathcal{U}^{\dagger}=\Lambda_G$ for $U=C^nZ_{m}$ with $n=1, m\geq 3$ or $n\geq 2, m\geq 2$. Proof is done.
\end{proof}

\subsection{Structure of $\mathsf{W}_X$}
In this part, we present the concrete form of $\mathsf{W}_X$. Following the discussion of $\mathsf{W}_X$ in previous subsections, we map $\mathsf{W}_X$ into $\mathbb{Z}_{m}^{2^{n+1}}$ via Eq.~\eqref{eq:Wxmap}.

We first discuss the cardinality of $\mathsf{W}_X$. The image
$\phi(\mathsf{W}_X)$ is the linear span of bases $\{v_0=e_0-e_{2^{n+1}-1},\cdots,v_{2^{n+1}-2}=e_{2^{n+1}-2}-e_{2^{n+1}-1}\}$. If considering the factor of global phase, the image would be the linear span of bases $\{v_0=e_0-e_{2^{n+1}-1},\cdots,v_{2^{n+1}-2}=e_{2^{n+1}-2}-e_{2^{n+1}-1}, u = 2^{\min(n+1, k)}e_{2^{n+1}-1}\}$ and we denote this enlarged image to be $\phi'(\mathsf{W}_X)$. In $\phi'(\mathsf{W}_X)$, any two vectors differing by $v_{2^{n+1}-1}=\begin{pmatrix}1&\cdots&1\end{pmatrix}^T$ correspond to the same gate. If $n+1\geq k$, $u$ is $\begin{pmatrix}0&\cdots&2^{k}\end{pmatrix}^T$. Now we consider the cardinality of $\phi'(\mathsf{W}_X)$. One can take arbitrary values from $\mathbb{Z}_m$ to fix the coefficients of the first $2^{n+1}-1$ bases. After that, the coefficient of the last basis $u$ only has $\frac{m}{2^k}$ inequivalent choices from $\mathbb{Z}_m$ as $(a+\frac{m}{2^k})u=au$. Hence, the cardinality of $\phi'(\mathsf{W}_X)$ is
\begin{equation}
\end{equation}
For any $\phi(w)$ belongs to $\phi'(\mathsf{W}_X)$, we can also find $\phi(w)+\begin{pmatrix}1&\cdots&1\end{pmatrix}^T,\cdots,\phi(w)+(m-1)\cdot\begin{pmatrix}1&\cdots&1\end{pmatrix}^T$ in $\phi'(\mathsf{W}_X)$. All these $m$ terms correspond to the same gate. Thus, the cardinality of $\phi(\mathsf{W}_X)$ is just $|\phi'(\mathsf{W}_X)|$ divided by $m$, excluding the redundacy of the global phase,
\begin{equation}\label{eq:cardWx}
\end{equation}

Now we investigate the structure of $\mathsf{W}_X=\langle\Pi^{\dagger} C^{n}Z_{m}\Pi C^{n}Z_{m}^{\dagger}\rangle$. We first analyze the simple case of $m=q$ and $m=2q$ where $q$ is an odd number. And then we use induction to analyze the general case that $m=q2^k$.

\begin{enumerate}[1)]
\item
Case 1: $m=q$.
\\
In this case,  $k = 0$, $\gcd(m, 2^{n+1})=1$, the vector $u=e_{2^{n+1}-1}$, and $\phi'(\mathsf{W}_X) = \mathbb{Z}_{m}^{2^{n+1}}$. Clearly, for any $0\leq l\leq n$, $\phi(C^l Z_m)\in \phi'(\mathsf{W}_X)$, so we can obtain that $\langle C^{n}Z_{m}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m} \rangle \leq \mathsf{W}_X$. Since $\langle C^{n}Z_{m}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m} \rangle = \langle C^{n}Z_{m}\rangle\times\langle C^{n-1}Z_{m}\rangle\times\cdots\times \langle Z_{m}\rangle$, $\abs{\langle C^{n}Z_{m}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m} \rangle} = m^{\binom{n+1}{n+1}}\times m^{\binom{n+1}{n}}\times\cdots\times m^{\binom{n+1}{1}}=m^{2^{n+1}-1} = \abs{\mathsf{W}_X}$. The cardinalities of $\mathsf{W}_X$ and its subgroup $\langle C^{n}Z_{m}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m} \rangle$ are the same, hence, we can conclude that
\begin{equation}
\mathsf{W}_X = \langle C^{n}Z_{m}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m} \rangle = \langle C^{n}Z_{m}\rangle\times\langle C^{n-1}Z_{m}\rangle\times\cdots\times \langle Z_{m}\rangle.
\end{equation}

\item\label{item:case2}
Case 2: $m=2q$.
\\
In this case, $k=1$, $\gcd(m, 2^{n+1})=2$, and $u=2e_{2^{n+1}-1}$. We can use the following formula to construct all $C^lZ_{m}$ with $0\leq l\leq n-1$,
\begin{equation}
C^lZ_{m}=\phi^{-1}\Bigl(v_{(2^{l+1}-1)*2^{n-l}}+v_{(2^{l+1}-1)*2^{n-l}+1}+\cdots+v_{2^{n+1}-2}+2^{n-l-1}\cdot u\Bigr).
\end{equation}
But this construction is not valid for $C^{n}Z_{m}$ as it requires $\frac{1}{2}u$. We can only construct $C^{n}Z_{m}^2=C^{n}Z_{m/2}$. This implies that $\langle C^{n}Z_{m/2},C^{n-1}Z_{m},C^{n-2}Z_{m},\cdots\rangle\leq \mathsf{W}_X$. Since $\langle C^{n}Z_{m/2},C^{n-1}Z_{m},\cdots\rangle=\langle C^{n}Z_{m/2}\rangle\times\langle C^{n-1}Z_{m}\rangle\times\cdots\times \langle Z_{m}\rangle$, $\abs{\langle C^{n}Z_{m/2}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m} \rangle} = (m/2)^{\binom{n+1}{n+1}}\times m^{\binom{n+1}{n}}\times\cdots\times m^{\binom{n+1}{1}}=m^{2^{n+1}-1}/2 = \abs{\mathsf{W}_X}$. Thus, in this case,
\begin{equation}
\mathsf{W}_X = \langle C^{n}Z_{m/2}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m} \rangle = \langle C^{n}Z_{m/2}\rangle\times\langle C^{n-1}Z_{m}\rangle\times\cdots\times \langle Z_{m}\rangle.
\end{equation}

\item
Case 3.1: $m=q2^k$ with $k\leq n$.
\\
We first give the result and prove it by induction.
\begin{theorem}
If $m=q2^k$ and $k\leq n$, the structure of $\mathsf{W}_X$ can be expressed in the following recursion,
\begin{equation}
\begin{split}
\mathsf{W}_X&=\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m}, \cdots, Z_m \rangle\times \langle A_k\rangle;\\
A_0&=\{\mathbb{I}\}, A_1=\{C^{n}Z_{m/2}\},\\ \forall 2\leq k\leq n, A_k&=\{C^{n-k+1} Z_{m/2}, gC^{n-k+1}Z_m| g\in A_{k-1} \}.
\end{split}
\end{equation}
\end{theorem}

\begin{proof}
The proof idea is showing $|\mathsf{W}_X| = |\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m}, \cdots, Z_m \rangle\times \langle A_k\rangle|$ and $\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m}, \cdots, Z_m \rangle\times \langle A_k\rangle \leq \mathsf{W}_X$.

Let us start with studying the relation between $\langle A_{k-1}\rangle$ and $\langle A_{k}\rangle$. For any $g\in \langle A_{k-1}\rangle$, it can be decomposed as $g=g_1^{\lambda_1}g_2^{\lambda_2}\cdots$ where $g_1,g_2,\cdots$ all belong to $A_{k-1}$ and $\lambda_1,\lambda_2,\cdots$ are integers. Since $g_1C^{n-k+1}Z_{m},g_2C^{n-k+1}Z_{m},\cdots\in A_k$, we can conclude that $g'=(g_1C^{n-k+1}Z_{m})^{\lambda_1}(g_2C^{n-k+1}Z_{m})^{\lambda_2}\cdots=gC^{n-k+1}Z_{m}^{\sum_{i}\lambda_i}\in\langle A_{k}\rangle$. Suppose $\sum_{i}\lambda_i$ is an even number, since $C^{n-k+1}Z_{m/2}=C^{n-k+1}Z_{m}^2\in A_{k}$, the set $\{g, gC^{n-k+1}Z_{m}^2, gC^{n-k+1}Z_{m}^4,\cdots\}\subseteq\langle A_k\rangle$. Now we want to argue that any gate has the form of $gC^{n-k+1}Z_{m}^{2p+1}$ cannot be in $\langle A_{k}\rangle$. This claim can be proved by contradiction. If $gC^{n-k+1}Z_{m}^{2p+1}\in\langle A_k\rangle$, combining with $gC^{n-k+1}Z_{m}^{2p}\in\langle A_k\rangle$, it is clear that $C^{n-k+1}Z_{m}=(gC^{n-k+1}Z_{m}^{2p})^{\dagger}gC^{n-k+1}Z_{m}^{2p+1}\in\langle A_k\rangle$. It leads to a contradiction as $C^{n-k+1}Z_{m}$ cannot be constructed with $\{v_i, 0\leq i\leq 2^{n+1}-2\}$ and $u = 2^ke_{2^{n+1}-1}$. Similarly, if $\sum_{i}\lambda_i$ is an odd number, we can see that all elements has the form $gC^{n-k+1}Z_{m}^{2p+1}$ will belong to $\langle A_k\rangle$ while elements within form of $gC^{n-k+1}Z_{m}^{2p}$ will not. In either case, one element $g$ in $\langle A_{k-1}\rangle$ will contribute to $|\langle C^{n-k+1}Z_{m/2}\rangle|=\frac{1}{2}|\langle C^{n-k+1}Z_{m}\rangle|$ elements in $\langle A_k\rangle$, which implies that $|\langle A_k\rangle|=\frac{1}{2}|\langle A_{k-1}\rangle|\times|\langle C^{n-k+1}Z_{m}\rangle|$. By iteration, the cardinality of $\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m},\cdots\rangle\times \langle A_k\rangle$ is
\begin{equation}\label{eq:cardequal}
\begin{split}
|\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m},\cdots\rangle\times \langle A_k\rangle|=&\frac{1}{2}|\langle C^{n-k+1}Z_{m}, C^{n-k}Z_{m},\cdots\rangle\times \langle A_{k-1}\rangle|\\
=&\frac{1}{2^k}|\langle C^{n}Z_{m}, C^{n-1}Z_{m},\cdots\rangle|\\
\end{split}
\end{equation}
This is exactly equal to $|\mathsf{W}_X|$. To complete the proof, we only need to argue that $\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m},\cdots\rangle\times \langle A_k\rangle\leq \mathsf{W}_X$. Note that $\phi'(\mathsf{W}_X)$ is spanned by bases ${v_i, 1\leq i\leq 2^{n+1}-2}$ and $u$. We only need to investigate whether the generators of $\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m},\cdots\rangle\times \langle A_k\rangle$ belong to the preimage set of $\phi$.

For $C^{n-k-l}Z_{m}$ with $0\leq l\leq n-k$, it can be expressed in the form of
\begin{equation}
\begin{split}
C^{n-k-l}Z_{m} &=\phi^{-1}\Bigl(2^{l}\begin{pmatrix}0&\cdots&2^{k}\end{pmatrix}^T+\sum_{i=(2^{n+1-k-l}-1)\cdot2^{k+l}}^{2^{n+1}-2}v_i\Bigr)\\
\end{split}
\end{equation}
Thus, $C^{n-k-l}Z_{m}$ belongs to $\mathsf{W}_X$. Next we prove $\langle A_k \rangle \leq \mathsf{W}_X$, or equivalently, $A_k \subseteq \mathsf{W}_X$, by induction. Denote $\mathsf{W}_X^k = \langle\Pi^{\dagger} C^{n}Z_{q2^k}\Pi C^{n}Z_{q2^k}^{\dagger}\rangle$. Suppose $A_{k-1}\subseteq \mathsf{W}_X^{k-1}$. Then, any element $g\in A_{k-1}$ has the following expression,
\begin{equation}
g=\phi^{-1}\Bigl(\begin{pmatrix}0&\cdots& 2^{k-1}\end{pmatrix}^T+\sum_{i=0}^{2^{n+1}-2}c_iv_i\Bigr).
\end{equation}
Note that $C^{n-k+1}Z_{m}$ can also be expressed in the above form as
\begin{equation}
C^{n-k+1}Z_{m}=\phi^{-1}\Bigl(\begin{pmatrix}0&\cdots&2^{k-1}\end{pmatrix}^T+\sum_{i=(2^{n-k+2}-1)\cdot 2^{k-1}}^{2^{n+1}-2}v_i\Bigr).
\end{equation}
This implies that $C^{n-k+1}Z_{m}^2$ and $C^{n-k+1}Z_{m}g$ can be expressed as below.
\begin{align}
C^{n-k+1}Z_{m}^2=&\phi^{-1}\Bigl(\begin{pmatrix}0&\cdots&2^{k}\end{pmatrix}^T+\sum_{i=(2^{n-k+2}-1)\cdot 2^{k-1}}^{2^{n+1}-2}2v_i\Bigr);\\
gC^{n-k+1}Z_{m}=&\phi^{-1}\Bigl(\begin{pmatrix}0&\cdots&2^{k}\end{pmatrix}^T+\sum_{i=0}^{(2^{n-k+2}-1)\cdot 2^{k-1}-1}c_iv_i+\sum_{i=(2^{n-k+2}-1)\cdot 2^{k-1}}^{2^{n+1}-2}(c_i+1)v_i\Bigr).
\end{align}
Therefore, any element $g'\in A_k$ has the expression,
\begin{equation}
g'=\phi^{-1}\Bigl(\begin{pmatrix}0&\cdots& 2^{k}\end{pmatrix}^T+\sum_{i=0}^{2^{n+1}-2}c_iv_i\Bigr).
\end{equation}
It means that all elements of $A_{k}$ can be constructed with $u$ and $v_i$, and belong to $\mathsf{W}_X^k$.
Now we have shown that $A_{k-1}\subseteq \mathsf{W}_X^{k-1}$ implies $A_{k}\subseteq \mathsf{W}^k_X$. For the induction argument to hold, we only need to check the initial condition, i.e., $A_1 \subseteq \mathsf{W}_X^{1}$. This reduces to Case~\ref{item:case2}. Thus, we prove that $A_k\subseteq \mathsf{W}_X$ and $\langle A_k\rangle\leq \mathsf{W}_X$. Hence, $\langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m},\cdots,Z_m\rangle\times \langle A_k\rangle\leq \mathsf{W}_X$. Combining with Eq.~\eqref{eq:cardequal}, we obtain that $\mathsf{W}_X = \langle C^{n-k}Z_{m}, C^{n-k-1}Z_{m},\cdots\rangle\times \langle A_k\rangle$ and proof is done.
\end{proof}

\item
Case 3.2: $m=q2^k$ with $k>n$.
\\
In this case, $\gcd(2^{n+1},m)=2^{n+1}$ and $\mathsf{W}_X=\langle A_{n+1}\rangle$ as described in the previous case.

Below we present examples of the twirling groups for target gate $U$ to be $C^nZ$ and $CZ_m$, which have been mentioned in the main text. In the next subsection, we will analyze their sample complexity and computational complexity.

\end{enumerate}

\begin{example}\label{exp:cnz}
The optimal twirling group $\mathsf{G}$ in CRU for multi-qubit controlled $Z$ gate $C^n Z$ is
\begin{equation}
\langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, Z, X \rangle = \mathsf{X}\ltimes \langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, Z\rangle.
\end{equation}
\end{example}

\begin{example}\label{exp:czm}
The optimal twirling group $\mathsf{G}$ in CRU for controlled phase gate $CZ_m$ is
\begin{equation}
\langle CZ_m, Z_m, X \rangle = \mathsf{X}\ltimes \langle CZ_m, Z_m\rangle,
\end{equation}
if $m$ is odd, and
\begin{equation}
\langle CZ_{m/2}, Z_m, X \rangle = \mathsf{X}\ltimes \langle CZ_{m/2}, Z_m\rangle,
\end{equation}
if $m$ is even and $m/2$ is odd, and
\begin{equation}
\langle CZ_{m/4}, Z_{m/2}, X, CZ_{m/2}Z^1_{m}, CZ_{m/2}Z^2_{m} \rangle = \mathsf{X}\ltimes \langle CZ_{m/4}, Z_{m/2}, CZ_{m/2} Z^1_{m}, CZ_{m/2}Z^2_{m} \rangle,
\end{equation}
if $m/2$ is even. It is worth mentioning that $\langle CZ_{m/4}, Z_{m/2}, CZ_{m/2} Z^1_{m}, CZ_{m/2}Z^2_{m} \rangle$ is a subgroup of $\langle CZ_{m/2}, Z_m\rangle$.
\end{example}

\subsection{Complexity analysis}\label{appendssc:complexity}
To enable randomized benchmarking or any other quantum information tasks with twirling groups, one needs to sample from the group and also needs to compute the multiplication of group elements. It is necessary to analyze the sample complexity and the computational complexity of the twirling group. The sample complexity is directly related to the cardinality of the corresponding twirling group. The computational complexity is related to both the group structure and the algorithm for computing the multiplication. We will provide a group multiplication algorithm and present its complexity as the upper bound of the group computational complexity. In the discussion below, we distinguish the number of qubits $N$ and the number of controlled qubits $n$. In general, $n\le N-1$.

To benchmark $C^{n}Z$ gate, the twirling group is $\mathsf{G}_{C^{n}Z} = \langle C^{n-1}Z,C^{n-2}Z\cdots,CZ,Z,X\rangle$ shown in Example~\ref{exp:cnz}. Note that this group has a semi-direct product structure,
\begin{equation}
\langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, Z, X\rangle=\langle X\rangle \rtimes(\langle C^{n-1}Z\rangle\times\langle C^{n-2}Z\rangle\times\cdots\langle Z\rangle).
\end{equation}
This means that any element in $\mathsf{G}_{C^{n}Z}$ can be written in the form of $\Pi W_{n-1}W_{n-2}\cdots W_{1}W_{0}$ where $\Pi\in\langle X\rangle$ and $W_{l}\in\langle C^{l}Z\rangle$. To sample from this twirling group $\mathsf{G}_{C^{n}Z}$, we just need to sample independently from $\langle C^{n-1}Z\rangle$, $\langle C^{n-2}Z\rangle$, $\cdots$, $\langle Z\rangle$, and $\langle X\rangle$. Since the orders of all generators are $2$, we can use a binary string whose length equals the number of generators of the group to represent an arbitrary group element. Then, we can sample the group element by sampling the binary string. For instance, the generators of $\langle CZ\rangle$ are $\{CZ_{1,2},CZ_{1,3},\cdots,CZ_{1,N},\cdots,CZ_{2,3},\cdots,CZ_{N-1,N}\}$. So a binary string with length $\binom{N}{2}$ is enough for sampling from $\langle CZ\rangle$. The total length needed to sample $\mathsf{G}_{C^{n}Z}$ is given by
\begin{equation}
\log |\mathsf{G}_{C^{n}Z}| = \log{\bigl|\langle X\rangle\rtimes\bigl(\langle C^{n-1}Z\rangle\times\cdots\times\langle CZ\rangle\times\langle Z\rangle\bigr)\bigr|}=n+\sum_{i=1}^{n}\binom{N}{i}=O(N^{n}),
\end{equation}
which is indeed the sample complexity. For computing the inverse gates, we will utilize the following identity. Given a $C^{n}Z$ gate acting on qubits $i_1,\cdots,i_{n+1}$ and a subset of $\{i_1,\cdots,i_{n+1}\}$, $I$, then
\begin{equation}\label{eq:swapidentity}
C^{n}Z_{i_1,i_2,\cdots,i_{n+1}}\Pi_{i\in I}X_i=\Pi_{i\in I}X_i\Pi_{S\subseteq I}C^{n-|S|}Z_{\{i_1,\cdots,i_{n+1}\}\backslash S}.
\end{equation}
%	C^{n}Z_{i_1,i_2,\cdots,i_n}X_{i_k}=X_{i_k}C^{n-1}Z_{i_1,\cdots,i_{k-1},i_{k+1},\cdots,i_{n}}\cdot C^{n}Z_{i_1,i_2,\cdots,i_n}.	\label{eq:C2}
So the multiplication of two elements in the group is calculated by
\begin{equation}
\begin{split}
&\Pi^{(1)}W_{n-1}^{(1)}\cdots W_{0}^{(1)} \Pi^{(2)}W_{n-1}^{(2)}\cdots W_{0}^{(2)}\\
=&(-1)^{\Pi^{(2)}\cdot W_{0}^{(1)}}\Pi^{(1)}W_{n-1}^{(1)}\cdots W_{1}^{(1)}\Pi^{(2)}W_{0}^{(1)}W_{n-1}^{(2)}\cdots W_{0}^{(2)}\\
=&(-1)^{\Pi^{(2)}\cdot W_{0}^{(1)}}\Pi^{(1)}W_{n-1}^{(1)}\cdots W_{2}^{(1)}\Pi^{(2)}W'_{0}W_{1}^{(1)}W_{0}^{(1)}W_{n-1}^{(2)}\cdots W_{0}^{(2)}\\
=&(-1)^{\Pi^{(2)}\cdot W_{0}^{(1)}}\Pi^{(1)}\Pi^{(2)} W'_{n-2}W_{n-1}^{(1)}\cdots W'_{0}W_{1}^{(1)}W_{0}^{(1)}W_{n-1}^{(2)}W_{n-2}^{(2)}\cdots W_{1}^{(2)}W_{0}^{(2)}.
\end{split}
\end{equation}
The $W'_0, \cdots, W'_{n-2}$ in the third and fourth lines are indeed the additional controlled-Z gates and multi-qubit controlled-Z gates in Eq.~\eqref{eq:swapidentity}.
%For certain ${W_{m}}'$, the calculation complexity is $O\Bigl(2^{m+1}\cdot\binom{N}{m+1}\Bigr)$.

Now we give a brief analysis of the total complexity. The key point is utilizing Eq.~\eqref{eq:swapidentity} to shift $\Pi^{(2)}$ gate to the left. We first fix an integer $l$ and focus on $C^{l}Z$. Now we try to study the swapping between $C^lZ$ and $\Pi^{(2)}$. By Eq.~\eqref{eq:swapidentity}, the total complexity for shifting $\Pi^{(2)}$ across a $C^{l}Z$ gate is $O(2^{l+1})$ since the cardinality of subset $I$ is at most $l+1$ and the number of subsets of $I$ is at most $2^{l+1}$. As the number of $C^{l}Z$ gates on $N$ qubits is at most $\binom{N}{l+1}$, the total complexity for shifting $\Pi^{(2)}$ to the leftmost is
\begin{equation}\label{eq:multiplycomplexity}
\text{group multiplication complexity}=\sum_{l=0}^{n-1}\binom{N}{l+1}\cdot 2^{l+1}.
\end{equation}
The right part is bounded by $N^{n}$. Hence the total group multiplication complexity is $O(N^{n})$. Notice that this bound is quite untight. For example if we let $n=N$, the right part of Eq.~\eqref{eq:multiplycomplexity} is indeed $3^{n}-1$.

%And for ${W_{m}}'$, it's computation comlexity is $O\bigl(n\cdot\binom{N}{m+1}\bigr)$ because we will need to use \ref{eq:C2} at most $n\cdot\binom{N}{m}$ times.
The inverse of an element $\Pi W_{n-1}\cdots W_{0}$ is $W_{n-1}\cdots W_{0}\Pi$, which equals the multiplication of element $W_{n-1}\cdots W_{1}1_{\langle X\rangle}$ and $1_{\langle C^{n-1}Z\rangle}\cdots 1_{\langle Z\rangle}\Pi$, where $1_{\mathsf{G}}$ represents the identity in group $\mathsf{G}$ and is indeed the all-zero string in the binary string representation of group elements. The computational complexity for computing inverse gate is then equal to the complexity for group element multiplication, i.e. $O(N^{n})$.

To benchmark $CZ_m$ gate, the twirling group by our protocol would always be a subgroup of $\mathsf{G}_{CZ_m}=\langle CZ_m, Z_m, X\rangle$. For simplicity, we only analyze the complexity of $\langle CZ_m, Z_m, X\rangle$. This complexity would be the upper bound of the complexity of the twirling group shown in Example~\ref{exp:czm}. $\langle CZ_m, Z_m, X\rangle$ has a similar direct product structure like $\langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, Z, X\rangle$,
\begin{equation}
\mathsf{G}_{CZ_{m}}=\langle X\rangle\rtimes(\langle CZ_m\rangle\times\langle Z_m\rangle).
\end{equation}
Thus, each element in $\mathsf{G}_{CZ_m}$ can be expressed as $W_2W_1\Pi$ where $W_2\in\langle CZ_m\rangle,W_1\in\langle Z_m\rangle$ and $\Pi\in\langle X\rangle$. In this case, the orders of generators $CZ_m$ and $Z_m$ are both $m$. We need to use $\lceil\log{m}\rceil$ bits to record their multiplicities and to uniquely identify one group element. Then, to represent an element in $\langle CZ_m\rangle$, we should use a bit string with length $\lceil\log{m}\rceil N(N-1)/2$. The total number of bits to express a group element is
\begin{equation}
\log\bigl|\langle X\rangle\rtimes(\langle CZ_m\rangle\times\langle Z_m\rangle)\bigr|=\left(\frac{N(N-1)}{2}+N\right)\lceil\log{m}\rceil+N=O(N^2\log{m}),
\end{equation}
which is indeed the sample complexity. The equation Eq.~\eqref{eq:swapidentity} can be extended for $CZ_m$ in the following way
\begin{equation}	CZ^{k}_{m}X_i=X_iCZ_{m}^{-k}(Z_{m})_j^{k},
\end{equation}
where the $CZ_{m}$ here is on qubit $i,j$. By similar argument like $C^nZ$, we can see that the group multiplication complexity, and thus inverse computation complexity, is $O(N^2\log m)$.
%assuming the $\log{m}$-bit basic operations can be done in $O(1)$ time.

The group size and computational complexity for the CNOT dihedral group have been elaborated in \cite{Cross2016dihedral}. So, we omit the details here. Note that for $CZ_m$, their results only apply to $m=2^k$, and ours admit $m$ taking arbitrary positive integers. The scaling of the complexity results has been summarized and listed in the main text. In Fig.~\ref{fig:card}, we also show an accurate result of the size comparison between our group and the CNOT dihedral group by considering $N = n + 1$. The size of our group increases slower with respect to the qubit number than that of the CNOT dihedral group.

\begin{figure}[htbp!]
\centering
\includegraphics[width=.48\textwidth]{./figure/GroupCompare.pdf}
\caption{Size comparison between the twirling groups for $C^nZ$ in our method and in~\cite{Cross2016dihedral}, associated with blue and orange patterns, respectively. $\abs{\textsf{G}}$ represents the group size. If taking a double logarithmic scale, both two kinds of groups increased nearly linearly with respect to the qubit number, but the CNOT dihedral group increases faster. The smaller figure demonstrates the results in the logarithm scale.}
\label{fig:card}
\end{figure}

%	\centering
%	\resizebox{\textwidth}{!}{ % 表格环境外部设置（头）
%		\begin{tabular}{|c|c|c|c|} \hline % 其中，|c|表示文本居中，文本两边有竖直表线。
%			& \multicolumn{3}{c|}{$C^nZ_{m}$}  \\ \hline
%			& Group & Sample Complexity & Computational Complexity \\ \hline
%			Our Work & $\langle C^{n}Z_{m}, C^{n-1}Z_{m}, \cdots, CZ_{m}, Z_{m}, X \rangle$ & $O(N^{n+1}\log{m})$ & $O(N^{n+1})$  \\ \hline
%			CNOT dihedral~\cite{Cross2016dihedral} & $\langle CX, Z_{2^{n+1}m}, X \rangle$ & $O((N^{n+1}+2^{N})\log{m})$ & $O(N^{3(n+1+\log{m})+1})$ when $m$ is a power of $2$ \\ \hline
%		\end{tabular}
%	}
%	\label{table:compelxity1}

Below, we discuss the hierarchy of twirling groups, which may be additionally interesting. In the main text, we have mentioned that by choosing the twirled gate to be $C^nZ_m$ with increasing $n$ and $m$, the size of the twirling group and the computational complexity of the group multiplication would keep increasing. The two complexities are closely related to the classical simulability of the twirling group, and the hierarchy of the groups actually forms a computational complexity hierarchy. The results may be useful in other fields like quantum complexity theory and quantum simulation.

Here, we discuss the hierarchy of groups themselves rather than the hierarchy of their sample and computational complexities. Specifically, we focus on the group of $\langle C^nZ, C^{n-2}Z, \cdots, CZ, X, Z_{2^k}\rangle$. Starting from Pauli group $\langle X, Z\rangle $, there are two ways to build up the hierarchy. One way is adding the number of sides of dihedral groups, i.e., increasing the number $k$ in $\langle X, Z_{2^k}\rangle$. The other way is adding multi-qubit controlled $Z$ gates $C^n Z$ or even adding multi-qubit controlled $X$ gates $C^n X$. Interestingly, $C^n Z$ can be contained in the CNOT-dihedral group $\langle CX, X, Z_{2^k}\rangle$ by choosing $k = n+1$~\cite{Cross2016dihedral}. The hierarchy is summarized below.
\begin{align}
\langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, X, Z_{2^k}\rangle &\leq \langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, X, Z_{2^{k+1}}\rangle \cdots \leq \langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, X, Z(\theta)\rangle \cdots;\\
\langle C^{n-1}X, X, Z_{2^k}\rangle &\leq \langle C^{n-1}X, X, Z_{2^{k+1}}\rangle \cdots \leq \langle C^{n-1}X, X, Z(\theta)\rangle \cdots \leq \mathrm{CRU}_n;\\
\langle X, Z_{2^k}\rangle &\leq \langle CZ, X, Z_{2^k}\rangle \leq \langle CCZ, CZ, X, Z_{2^k}\rangle \cdots;\\
\langle X, Z_{2^k}\rangle &\leq \langle CX, X, Z_{2^k}\rangle \leq \langle CCX, X, Z_{2^k}\rangle \cdots \leq \mathrm{CRU};\\
\langle C^{k-1}Z, C^{k-2}Z, \cdots, CZ, X, Z_{2^k}\rangle &\leq \langle CX, X, Z_{2^k}\rangle.
\end{align}
Here, $\mathrm{CRU}_n$ represents the CRU on $n$ qubits and $\mathrm{CRU}$ represents $\cup_{n} \mathrm{CRU}_n$. As shown in the above equations, the upper limit of the hierarchy is CRU. The key point is that $\langle C^{n-1}X, X\rangle$ contains all permutations over the computational bases $\{0,1\}^n$ based on Lemma~\ref{lemma:permutation}. Thus, $\langle C^{n-1}X, X, Z(\theta)\rangle$ contains all quantum gates like $\Pi W$ where $\Pi$ is an arbitrary permutation gate and $W$ is an arbitrary diagonal gate. This is exactly the necessary and sufficient condition for CRU. Also, from the group hierarchy, it is clearer that the twirling group of our protocol is smaller than that in~\cite{Cross2016dihedral} for multi-qubit controlled-$Z$ gates $C^nZ$. It is straightforward by investigating $\langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, X, Z\rangle \leq \langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, X, Z_{2^k}\rangle \leq \langle CX, X, Z_{2^k}\rangle$.

%Below we only consider the part for benchmarking $C^nZ_m$ in the hierarchy, which only involves groups $\langle C^{n-1}Z, C^{n-2}Z, \cdots, CZ, X, Z_m\rangle$ with increasing $n$ and $m$. To enable randomized benchmarking for $C^nZ_m$, we should study its sample complexity and computational complexity so that we can efficiently sample gates from it and compute and inverse gate in the quantum circuit. Below in Table~\ref{table:compelxity} we provide the scaling of the complexity and compare the results with that of~\cite{Cross2016dihedral}.

\section{Random compiling}\label{appendsc:RC}
In this part, we discuss the application of our results in random compiling. Recall that in the task of random compiling, we need to turn the noisy quantum gate $\widetilde{\mathcal{U}} = \mathcal{U} \Lambda$ to $\mathcal{U} \Lambda_G$, where the twirled noise channel $\Lambda_G$ is supposed to be a Pauli channel. Here, we consider a more general case than that in the main text. Assuming we implement a quantum circuit, $\widetilde{C} = \cdots \widetilde{\mathcal{U}}_2 \widetilde{\mathcal{U}}_1 = \cdots \mathcal{U}_2 \Lambda_2 \mathcal{U}_1 \Lambda_1$, as shown in Fig.~\ref{fig:RCappend}(a), the task of random compiling is tailoring $\widetilde{C}$ into $ \cdots \mathcal{U}_2 \Lambda_{2G} \mathcal{U}_1 \Lambda_{1G}$ as shown in Fig.~\ref{fig:RBappend}(c). The quantum gates $U_i, i\in \mathbb{Z}_+$ are tailored gates and, in general, are very noisy in experiments. They are always non-local, acting on several qubits. To tailor them, we add twirling gates $G_i, G_i'$ beside them as shown in Fig.~\ref{fig:RCappend}(b). The twirling gates are always noiseless and more easily implementable than the tailored gates.

\begin{figure}[htbp!]
\centering
\includegraphics[width=.5\textwidth]{./figure/RCappend.pdf}
\caption{Random compiling for $U_1$, $U_2$, $\cdots$, by inserting twirling gates $G_i$ and $G'_i = U_iG_i^{-1}U_i^{-1}$ beside $U_i$ with $i\in \mathbb{Z}_+$. $\Lambda_{iG}$ is the $\mathsf{G}$-twirled noise channel required to be a Pauli channel. In reality, $G_i$ and $G'_{i-1}$ are merged and implemented together as quantum gate $G_iG'_{i-1}$.}
\label{fig:RCappend}
\end{figure}

In the original work of~\cite{Wallman2016compiling}, the tailored multi-qubit quantum gates are restricted to be Clifford gates, and the twirling gates are chosen as Pauli gates. Nonetheless, in experiments, some multi-qubit non-Clifford gates, like CS and CCZ, are also important and native. They can be directly realized without gate composition. A random compiling scheme for these non-Clifford gates is conducive to their applications in quantum information tasks. Below, we show that our twirling groups constructed for benchmarking target gate $U$ can also be used to tailor $U$ in the setting of random compiling.

Similar to the discussion in RB, we investigate the requirements of the twirling gates for tailoring a gate set $\mathsf{U} = \{U_1, U_2, \cdots\}$. Note that in Figs.~\ref{fig:RCappend}(a),~\ref{fig:RCappend}(b), and~\ref{fig:RCappend}(c) we have shown the process to tailor $\cdots \mathcal{U}_2 \Lambda_2 \mathcal{U}_1 \Lambda_1$ into $ \cdots \mathcal{U}_2 \Lambda_{2G} \mathcal{U}_1 \Lambda_{1G}$. The key point is inserting twirling gates $G_i$ and $G'_i=U_iG_i^{-1}U_i^{-1}$ beside $U_i$ where $G_i$ is randomly sampled from twirling group $\mathsf{G}$. In reality, the gates $G_i$ and $G'_{i-1}$ are merged and implemented together to reduce quantum resource consumption. Thus, the twirling gates in general belong to $\mathsf{V} = \bigcup_{U\in \mathsf{U}} \mathsf{G}U\mathsf{G}U^{\dagger}$. Thus, in random compiling to tailor gate set $\mathsf{U}$, we should optimize the choice of $\mathsf{G}$ to simplify $\mathsf{V}$ as much as possible. It is equivalent to solving the following question.

\begin{question}\label{ques:RC}
Given a gate set, $\mathsf{U} = \{U_1, U_2, \cdots\}$, minimize $\abs{\mathsf{V}}$, such that $\mathsf{V} = \bigcup_{U\in \mathsf{U}} \mathsf{G}U\mathsf{G}U^{\dagger}$ and the twirling group $\mathsf{G}$ satisfies
\begin{equation}
\Lambda_G = \mathbb{E}_{G\in \mathsf{G}} \mathcal{G} \Lambda \mathcal{G}^{\dagger} \ \text{is a Pauli channel.}
\end{equation}
\end{question}

It can be seen that if the gate set $\mathsf{U}$ only contains a single gate $U$, then the solution to Question~\ref{ques:RB} is at least a feasible solution to Question~\ref{ques:RC}. The solution may not be optimal but will provide an upper bound on $\abs{\mathsf{V}}$. Now we suppose that the gate set $\mathsf{U}$ only contains a multi-qubit controlled phase gate $U=C^nZ_m$. Question~\ref{ques:RC} reduces to finding the smallest set of $\mathsf{G}U\mathsf{G} U^{\dagger}$ while $\Lambda_G$ is a Pauli channel. Similar with the results in RB, if we assume $\mathsf{G}$ belonging to CRU, then $\mathsf{G}U \mathsf{G} U^{\dagger}$ cannot be smaller than $\mathsf{X}U\mathsf{X}U^{\dagger}$ for $U=C^nZ_m$ based on Lemma~\ref{lemma:cosetcardconstraint}. Meanwhile, by choosing $\mathsf{G}$ as Pauli group $\mathsf{P}_n$, we would obtain a solution $\mathsf{V}$ to Question~\ref{ques:RC}, which is $\mathsf{P}_nU\mathsf{P}_n U^{\dagger} = \mathsf{X}U\mathsf{X}U^{\dagger} \mathsf{Z}$. Note that $\log |\mathsf{V}| - \log |\mathsf{X}U\mathsf{X}U^{\dagger}| \leq \log |\mathsf{Z}| = n$. It means that the size of $\mathsf{P}_nU\mathsf{P}_n U^{\dagger}$ would not be much larger than the optimal solution since the optimal solution contains $\mathsf{X}U\mathsf{X}U^{\dagger}$. Then, we found the nearly optimal solution for tailoring a multi-qubit controlled phase gate, $C^nZ_m$, in the setting of random compiling. It is simpler than the twirling group associated with $C^nZ_m$ in the setting of RB. Generally speaking, the requirements for the twirling gates in random compiling are much simpler than those in RB. We leave the problem of tailoring more kinds of quantum gates in the setting of random compiling to the future.

\section{Simulation}\label{appendsc:simulation}
In this section, we present the details of the simulation, including the setting of the noise model, the benchmarking protocol using the proposed twirling groups, and an additional benchmarking result for CS gate, which equals $\begin{pmatrix}
1  & 0\\
0 & i
\end{pmatrix}$. We denote the total number of qubits as $N$, which is set to be $2$ for the CS gate and $n+1$ for the $C^{n}Z$ gate.

\subsection{Error model}
In our simulation, we assume the twirling gates are always noiseless and only consider the noise of the target gate since the noise of twirling groups is easy to measure and eliminate by various methods that we mentioned before.
We generate this noise channel by composing three kinds of noise channels:
\begin{enumerate}[1.]
\item
Depolarize channel $\Lambda_d(\rho) = p\rho + (1-p)\frac{\mathbb{I}}{2^N}$, in which $\mathbb{I}$ represents the identity operator with dimension $2^N\times 2^N$. In our simulation experiments, $p$ is prefixed to be a constant.

\item
Local amplitude damping channel $\Lambda_a$. The amplitude damping channel on qubit $i$, where $1\leq i\leq N$, is defined as
\begin{equation}
\Lambda_a^i(\rho) = K_0^i\rho K_0^{i\dagger} + K_1^i\rho K_1^{i\dagger}.
\end{equation}
\begin{equation}
K_0^i \equiv \begin{pmatrix}
1 &  0\\
0 & \sqrt{p_i}
\end{pmatrix},
K_1^i \equiv \begin{pmatrix}
0 & \sqrt{1-p_i} \\
0 & 0
\end{pmatrix}.
\end{equation}
The parameter $p_i$ denotes the noise strength of the amplitude damping channel and is uniformly and randomly sampled from $[0, 0.02]$. Then, the local amplitude damping noise on $N$ qubits is defined as $\Lambda_a = \bigotimes_{i=1}^N \Lambda_a^i$.

\item
Unitary channel $\Lambda_u$. We consider that there exist two types of unitary noise for the target gate at the same time. One is SWAP-coupling noise, and the other is multi-Z-coupling noise. The SWAP-coupling noise is defined as
\begin{equation}
U_{\text{SWAP}}\equiv e^{i \sum_{1 \leq j < k \leq N} \alpha_{jk}\text{SWAP}_{jk}},
\end{equation}
where $\text{SWAP}_{jk}$ denotes the SWAP gate between qubits $j$ and $k$. The parameter $\alpha_{jk}$ is used to adjust the noise strength of SWAP-coupling noise and is uniformly and randomly sampled from $[0, 0.01]$. The ZZ-coupling noise is defined as
\begin{equation}
U_{\text{ZZ}}\equiv e^{i \sum_{z\in \{0,1\}^N, \abs{z}\geq 2} \beta_z \ketbra{z}}.
\end{equation}
The parameter $\beta_{z}$ represents the coupling strength and is uniformly and randomly sampled from $[0, 0.1]$. This noise records the phase fluctuation of the computational basis with a weight of no less than 2. The total unitary noise is set as $U = U_{\text{ZZ}}U_{\text{SWAP}}$, and the corresponding channel representation is $\Lambda_u$.
\end{enumerate}
Based on the three kinds of noise, the noise of the target gate is generated by $\Lambda = \Lambda_u \circ \Lambda_a \circ \Lambda_d$.

Besides the noise of the gate, we also set errors for state preparation and measurement (SPAM). In our simulation experiments, the initial states only contain $\ket{0}^{\otimes N}$ and $\ket{+}^{\otimes N} = H^{\otimes N} \ket{0}^{\otimes N}$ where $H$ is the Hadamard gate. In the simulation, we suppose the Hadamard gate is noiseless, and the error of the initial state only comes from the noise when preparing $\ket{0}^{\otimes n}$. We assume that on each qubit, the state $\ket{0}$ would turn into $\ket{1}$ with probability 0.02, resulting from the thermal excitation. We take the measurement to be noiseless as there is always a measurement error correction in experiments.

\subsection{Benchmarking protocol and additional simulation results}
Below, we present the benchmarking protocol for CCZ and CS with the twirling group $\text{CZD} = \langle CZ, X, S\rangle$. The benchmarking protocol for $C^nZ$ is almost the same except for substituting twirling group $\langle CZ, X, S\rangle$ with $\langle C^{n-1}, \cdots, CZ, X, S\rangle$.

To help readers more deeply understand the choice of the twirling groups and the benchmarking protocols, we take CS gate for an example and present the Liouville representation of its twirled noise channel under several different twirling groups, including Pauli group $\mathsf{P} = \langle X, Z \rangle$, CZ Pauli group $\text{CZP} = \langle CZ, X, Z \rangle$, CZ dihedral group $\text{CZD} = \langle CZ, X, S \rangle$, and CNOT dihedral group $\text{CXD} = \langle CX, X, T \rangle$. The results are shown in Figs.~\ref{fig:channelp},~\ref{fig:channelczp},~\ref{fig:channelcz}, and~\ref{fig:channelcx}.

\begin{figure}[htbp!]
\centering
\includegraphics[width=.85\textwidth]{./figure/channelp.pdf}
\caption{Pauli Liouville representation of $\Lambda_{\mathsf{P}}$ where $\Lambda$ is the noise channel of CS gate and $\mathsf{P}$ denotes the Pauli group. The values in vacant squares are all 0 and we omit them. The two axes record the bases of the Pauli-Liouville representation.}
\label{fig:channelp}
\end{figure}

\begin{figure}[htbp!]
\centering
\includegraphics[width=.85\textwidth]{./figure/channelczp.pdf}
\caption{Pauli Liouville representation of $\Lambda_{\textsf{CZP}}$ where $\Lambda$ is the noise channel of CS gate and \textsf{CZP} denotes the twirling group $\langle CZ, Z, X\rangle$. The values in vacant squares are all 0, and we omit them. Note that the elements corresponding to $X_1X_2$ and $Y_1Y_2$ are the same, and the elements corresponding to $X_1Y_2$ and $X_1Y_2$ are the same. But the two values are different. Generally, the elements corresponding to $X_1$ and $Y_1$ are different, and the elements corresponding to $X_2$ and $Y_2$ are also different. However, in this case, the noise channel is very special, and the difference can hardly be seen by accident.}
\label{fig:channelczp}
\end{figure}

\begin{figure}[htbp!]
\centering
\includegraphics[width=.85\textwidth]{./figure/channelcz.pdf}
\caption{Pauli Liouville representation of $\Lambda_{\textsf{CZD}}$ where $\Lambda$ is the noise channel of CS gate and \textsf{CZD} denotes the twirling group $\langle CZ, Z, S\rangle$. The values in vacant squares are all 0, and we omit them. The matrix elements corresponding to $X_a\mathsf{Z}$ are all averaged respectively for $X_a\in \{X_1, X_2, X_1X_2\}$.}
\label{fig:channelcz}
\end{figure}

\begin{figure}[htbp!]
\centering
\includegraphics[width=.85\textwidth]{./figure/channelcx.pdf}
\caption{Pauli Liouville representation of $\Lambda_{\textsf{CXD}}$ where $\Lambda$ is the noise channel of CS gate and \textsf{CXD} denotes the twirling group $\langle CX, X, T\rangle$. The values in vacant squares are all 0, and we omit them. The second to the fourth diagonal elements are averaged, and the last 12 diagonal elements are also averaged.}
\label{fig:channelcx}
\end{figure}

The matrix elements of the original noise channel $\Lambda$ of $CS$ gate in the Pauli-Liouville representation are all non-zero and contain complex numbers. After twirled by the Pauli group, there are only diagonal terms left to be nonzero, as shown in Fig.~\ref{fig:channelp}. If we use a larger twirling group, the CZ Pauli group, for any $X_a\in \{X_1, X_2, X_1X_2\}$, the matrix elements corresponding to $X_a$ and $X_aZ_1Z_2$ would be averaged, and the elements of $X_aZ_1$ and $X_aZ_2$ would be also averaged. The matrix elements of $\Lambda_{\textsf{CZP}}$ are shown in Fig.~\ref{fig:channelczp}. Its diagonal blocks are associated with the bases set in $\{ \{\mathbb{I}\}, \{Z_1\}, \{Z_2\}, \{Z_1Z_2\}, \{X_aZ_1, X_aZ_2\}, \{X_a, X_aZ_1Z_2\} | X_a\in \{X_1, X_2, X_1X_2\}\}$. Unfortunately, this group does not satisfy $\Lambda_{\textsf{CZP}}\mathcal{U} = \mathcal{U}\Lambda_{\textsf{CZP}}$ where $U$ is the target gate CS. The Pauli Liouville representation of the CS gate is shown in Fig.~\ref{fig:mapcs}. The diagonal blocks of the CS gate are associated with the bases set in $\{ \{\mathbb{I}\}, \{Z_1\}, \{Z_2\}, \{Z_1Z_2\}, \{X_a, X_aZ_1, X_aZ_2, X_aZ_1Z_2\} | X_a\in \{X_1, X_2, X_1X_2\}\}$ and are inconsistent with that of $\Lambda_{\textsf{CZP}}$. If we further use a bit larger twirling group, CZ dihedral group, the matrix elements corresponding to $X_a\mathsf{Z}$ would be all averaged. Thus, the diagonal blocks of $\Lambda_{\textsf{CZD}}$ are consistent with that of $\mathcal{U}$ and $\Lambda_{\textsf{CZD}}\mathcal{U} = \mathcal{U}\Lambda_{\textsf{CZD}}$ holds. Using the CZ dihedral group for twirling, we can obtain $\Lambda_{\textsf{CZD}}^m$. Combining $Z$-basis and $X$-basis measurements, we can extract the diagonal terms of $\Lambda_{\textsf{CZD}}^m$ and hence obtain the diagonal terms of  $\Lambda_{\textsf{CZD}}$ via fitting. The full protocol is shown in detail in Box~\ref{box:CnZmRB}. The situation for the CCZ gate is similar. Although CZ Pauli group $\langle CZ, Z, X\rangle$ can make $\Lambda_{\textsf{CZP}}\mathcal{U} = \mathcal{U}\Lambda_{\textsf{CZP}}$ for $U=\text{CCZ}$, the number of different parameters of $\Lambda_{\textsf{CZP}}$ is $3*(2^N-1)$ and is more than $2*(2^N-1)$. The parameters cannot be extracted only with $Z$-basis and $X$-basis measurements. On the other hand, the number of parameters of $\Lambda_{\textsf{CZD}}$ is equal to $2*(2^N-1)$. The parameters can be directly extracted only with $Z$-basis and $X$-basis measurements. Hence, we utilize the CZ dihedral group for tailoring CCZ to simplify the SPAM settings.\\

\begin{mybox}{box:CnZmRB}{Procedures for benchmarking CCZ or CS with CZ dihedral group}
\begin{enumerate}
\item
First initialize the state, $\ket{\psi_0} = \ket{0}^{\otimes N}$ or $\ket{\psi_1} = \ket{+}^{\otimes N}$ where $N$ is the number of qubits. $N=2$ for CS gate and $N=3$ for CCZ gate.

\item
Choose a positive integer, $M$, and choose two sets of positive integers $\{m_1, m_2, ..., m_M\}$ and $\{K_1, K_2, ..., K_M\}$. Here, $\{m_1, m_2, ..., m_M\}$ is the set of circuit depths and for $1\leq i\leq m$, $K_i$ is the number of sampled sequences when circuit depth equals $m_i$.

\item
For each integer $1\leq i\leq M$, uniformly and randomly sample $2m_i$ gates from $\langle CZ, S, X\rangle$ for $K_i$ times, which we denote $\{G_{j,1}, G_{j,2}, \cdots, G_{j,2m_i}\}$, $1\leq j\leq K_i$.

\item
For each integer $1\leq i\leq M$ and $1\leq j\leq K_i$, implement gate sequence
\begin{equation}
\widetilde{S}(j, m_i) = \widetilde{U}_{inv}\prod_{t=1}^{m_i} U^{\dagger}G_{j, 2t}UG_{j, 2t-1},
\end{equation}
where $\widetilde{\cdot}$ represents the noisy version of the quantum gate and $U_{inv} = (\prod_{t=1}^{m_i} U^{\dagger}G_{j, 2t}UG_{j, 2t-1})^{-1}$. $U$ is the target gate CCZ or CS.

\item
For initial state $\ket{\psi_0} = \ket{0}^{\otimes N}$, measuring all observables from $\{\mathbb{I}, Z\}^{\otimes N}$ of the final state via $Z$-basis measurement. For initial state $\ket{\psi_1} = \ket{+}^{\otimes N}$, measuring all observables from $\{\mathbb{I}, X\}^{\otimes N}$ of the final state via $X$-basis measurement. That is, for each $Q_k \in \{\mathbb{I}, Z\}^{\otimes N}$, estimate
\begin{equation}
f_{Z}(j, m_i, k) = \tr(\widetilde{Q}_k \widetilde{S}(j, m_i)(\widetilde{\rho}_0)).
\end{equation}
For each $P_k \in \{\mathbb{I}, X\}^{\otimes N}$, estimate
\begin{equation}
f_{X}(j, m_i, k) = \tr(\widetilde{P}_k \widetilde{S}(j, m_i)(\widetilde{\rho}_1)).
\end{equation}
Here, $\widetilde{\rho}_0$ and $\widetilde{\rho}_1$ are noisy versions of $\ket{0}^{\otimes N}$ and $\ket{+}^{\otimes N}$, respectively.

\item
Average the results of different gate sequences and obtain
\begin{align}
f_Z(m_i, k) &= \frac{1}{K_i}\sum_{j=1}^{K_i}f_{Z}(j, m_i, k)\\
f_X(m_i, k) &= \frac{1}{K_i}\sum_{j=1}^{K_i}f_{X}(j, m_i, k).
\end{align}

\item
For each $Q_k$, fit $f_Z(m_i, k)$ to function
\begin{equation}
f(m_i) = A\lambda_{Z,k}^{m_i}
\end{equation}
and obtain $\lambda_{Z,k}$. For each $P_k$, fit $f_X(m_i, k)$ to function
\begin{equation}
f(m_i) = A\lambda_{X,k}^{m_i}
\end{equation}
and obtain $\lambda_{X,k}$.

\item
Estimate the fidelity of the target gate via
\begin{equation}
F = \frac{\sum_k\lambda_{Z,k} + 2^N(\sum_k\lambda_{X,k} - 1)}{4^N}.
\end{equation}

\item
If one wants to separate further the noise of the twirling gates and the target gate, execute the following step. Replace the target gate with identity and repeat the above processes to estimate the fidelity of the twirling groups, $F_G$. Then, estimate the fidelity of $U$ with $F$ and $F_G$ by the technique of interleaved RB~\cite{Magesan2012interleavedRB}. In our simulation, we omit this step by assuming the twirling gates are noiseless. This assumption is also taken for protocols using the CNOT dihedral group.
\end{enumerate}
\end{mybox}

Note that in the above benchmarking procedure, we use the circuit structure in~\cite{Zhang2023Scalable} instead of~\cite{Erhard2019CB}. That means we implement random twirling gates interleaved with $U$ and $U^{\dagger}$ instead of being just interleaved with $U$. For $C^nZ$ gate, $U=U_{inv}$ and this modification does not influence anything. For more general case that $U$ is $C^nZ_m$ gate like CS gate, under a mild assumption that $U$ and $U^{\dagger}$ has the same noise $\Lambda$, the above procedure can estimate the fidelity of $\sqrt{\mathcal{U}^{\dagger}\Lambda_G\mathcal{U}\Lambda_G}$, which is equal to the fidelity of $\Lambda$. In this modified procedure, the circuit depth only needs to be chosen as the multiples of 2. While within the circuit structure in Fig.~\ref{fig:RBappend}, just like~\cite{Erhard2019CB}, the circuit depth needs to be chosen as the multiples of the order of the target gate to ensure the inverse gate belonging to the twirling group. For instance, the order of $C^nZ_m$ is $m$ and can be large for $C^nZ_m$ with large $m$. In this case, the modified benchmarking procedure in Box~\ref{box:CnZmRB} can offer an advantage of shorter circuit depth in circuit implementation. One can also use the circuit structure in Fig.~\ref{fig:RBappend} to benchmark the channel.

If using CNOT dihedral group for twirling, the twirled noise channel $\Lambda_{\textsf{CXD}}$ would be more symmetric than $\Lambda_{\textsf{CZD}}$ and $\Lambda_{\textsf{CXD}}\mathcal{U} = \mathcal{U}\Lambda_{\textsf{CXD}}$ also holds. The elements corresponding to $Z_1$, $Z_2$, and $Z_1Z_2$ will be averaged and the elements corresponding to $\{X_a\mathsf{Z}| X_a\in \{X_1, X_2, X_1X_2\}\}$ will also be averaged. Thus, $\Lambda_{\textsf{CXD}}$ only has two parameters as shown in Fig.~\ref{fig:channelcx}. With the protocol in~\cite{Cross2016dihedral}, one can obtain these two parameters. Note that in~\cite{Cross2016dihedral}, there are two ways to extract the two parameters. One way is extracting the two parameters with double exponential fitting, which performs badly, and we do not use it. We simulate the other way that using two SPAM settings extracts the parameters:
\begin{enumerate}[a.]
\item \label{item:sma}
preparing and measuring $\ket{0}^{\otimes N}$;
\item \label{item:smb}
preparing and measuring $\ket{+}^{\otimes N}$,
\end{enumerate}
respectively. Within the first SPAM setting, one can nearly get $\bra{0}^{\otimes N}\Lambda^m_{\textsf{CXD}}\ket{0}^{\otimes N} \approx A\lambda_Z^m+B$ and obtain $\lambda_Z = \frac{\sum_k \lambda_{Z,k}-1}{2^N-1}$. Here, $\lambda_{Z,k}$ is the diagonal term of $\Lambda_{\textsf{CZD}}$ mentioned in Box~\ref{box:CnZmRB}. Similarly, with the second SPAM setting, one can nearly get $\bra{+}^{\otimes N}\Lambda^m_{\textsf{CXD}}\ket{+}^{\otimes N} \approx A\lambda_X^m+B$ and obtain $\lambda_X = \frac{\sum_k \lambda_{X,k}-1}{2^N-1}$. One can also use the state preparation and measurement settings and the post-processing methods in Box~\ref{box:CnZmRB} to obtain these two parameters.

Below, we show the simulation result of CS gate as a supplement. The result is similar to that of CCZ and $C^nZ$ gates in the main text, as shown in Fig.~\ref{fig:CS}. The result indicates that the benchmarking protocol with the CZ dihedral group can perform as well as that with the CNOT dihedral group if using the procedure in Box~\ref{box:CnZmRB}. Both two are better than the protocol using the CNOT dihedral group using SPAM settings~\ref{item:sma} and~\ref{item:smb}. It is reasonable as the protocol in Box~\ref{box:CnZmRB} utilizes all the bases of the measurements while settings~\ref{item:sma} and~\ref{item:smb} are only involved with one basis in one measurement.

\begin{figure}[htbp!]
\centering
\includegraphics[width=.5\textwidth]{./figure/CSboxlot.pdf}
\caption{Benchmarking results for CS with CZ dihedral group $\langle CZ, X, S\rangle$ and CNOT dihedral group $\langle CX, X, T\rangle$. There are two processing procedures for the CNOT dihedral group. The original procedure in~\cite{Cross2016dihedral}, or the procedure with SPAM settings~\ref{item:sma} and~\ref{item:smb}, is denoted as CXDn. The procedure using settings in Box~\ref{box:CnZmRB} is denoted as CXDt. The green dashed line is the theoretical value of the noise channel fidelity, and the other three box plots are the estimated fidelities of 100 groups. Each group employs circuits with depth list $\{m_1, m_2, ..., m_M\} = \{3, 6,\cdots, 30\}$. For each depth, we take the same number of sampled sequences, which means $K_1 = K_2 = ... = K_M$. The horizontal axis gives the value of $K_i, 1\leq i\leq M$.}
\label{fig:CS}
\end{figure}

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