u_{jl}=\sum_k u_{ik}=1$, we have: $$\sum_k \delta_{{\varepsilon}_{kl}=1}u_{ik}
=\sum_{k,j}\delta_{{\varepsilon}_{kl}=1}\delta_{{\varepsilon}_{ij}=1}u_{ik}u_{jl}
=\sum_{j}\delta_{{\varepsilon}_{ij}=1}u_{jl}$$ Conversely, assume that (iii) holds. We thus have: $$\sum_{k'} \delta_{{\varepsilon}_{k'l}=0}u_{ik'}=1-\sum_{k'} \delta_{{\varepsilon}_{k'l}=1}u_{ik'}
=1- \sum_{j'} \delta_{{\varepsilon}_{ij'}=1}u_{j'l}=\sum_{j'} \delta_{{\varepsilon}_{ij'}=0}u_{j'l}$$ Furthermore, ${\varepsilon}_{kl}=0$ and ${\varepsilon}_{k'l}=1$ implies $k\neq k'$ and hence $u_{ik}u_{ik'}=0$. Likewise we see $\delta_{{\varepsilon}_{ij'}=0}\delta_{{\varepsilon}_{ij}=1}u_{j'l}u_{jl}=0$. Therefore: $$\begin{aligned}
\delta_{{\varepsilon}_{kl}=0}u_{ik}u_{jl}
&=\sum_{k'}\delta_{{\varepsilon}_{kl}=0}\delta_{{\varepsilon}_{k'l}=0}u_{ik}u_{ik'}u_{jl}
+\sum_{k'}\delta_{{\varepsilon}_{kl}=0}\delta_{{\varepsilon}_{k'l}=1}u_{ik}u_{ik'}u_{jl}\\
&=\sum_{k'}\delta_{{\varepsilon}_{kl}=0}\delta_{{\varepsilon}_{k'l}=0}u_{ik}u_{ik'}u_{jl}\\
&=\sum_{j'} \delta_{{\varepsilon}_{kl}=0}\delta_{{\varepsilon}_{ij'}=0}u_{ik}u_{j'l}u_{jl}\\
&=\sum_{j'} \delta_{{\varepsilon}_{kl}=0}\delta_{{\varepsilon}_{ij'}=0}\delta_{{\varepsilon}_{ij}=0}u_{ik}u_{j'l}u_{jl}\end{aligned}$$ On the other hand: $$\begin{aligned}
\delta_{{\varepsilon}_{ij}=0}u_{ik}u_{jl}
&=\sum_{j'}\delta_{{\varepsilon}_{ij}=0}\delta_{{\varepsilon}_{ij'}=0}u_{ik}u_{j'l}u_{jl}\\
&=\sum_{k'}\delta_{{\varepsilon}_{ij}=0}\delta_{{\varepsilon}_{k'l}=0}u_{ik}u_{ik'}u_{jl}\\
&=\sum_{k'}\delta_{{\varepsilon}_{ij}=0}\delta_{{\varepsilon}_{k'l}=0}\delta_{{\varepsilon}_{kl}=0}u_{ik}u_{ik'}u_{jl}\end{aligned}$$ This proves that (ii) holds.

The previous lemma and the next proposition comparing the two different definitions of quantum automorphism groups of [@BicQAG] and [@BanQAG] may also be found in [@QAGThesis Sect. 3.1].

\[PropVglBB\] The ${\varepsilon}$-symmetric quantum group $S_n^{\varepsilon}$ coincides with the quantum automorphism group of $\Gamma_{\varepsilon}$ as defined by Bichon, and it is a quantum subgroup of the quantum automorphism group $S_n^{\Gamma_{\varepsilon}}$ as defined by Banica.

The relations (3.2) of Theorem 3.2 in [@BicQAG] are equivalent to ($\mathring R^{\varepsilon}2$) and his relations (3.3) are equivalent to ($\mathring R^{\varepsilon}1$). His relations (3.4) follow from (3.2) and (3.3) using for ${\varepsilon}_{kl}=1$: $$\sum_{ij}\delta_{{\varepsilon}_{ij}=1}u_{ik}u_{jl}=
\sum_{ij}u_{ik}u_{jl}
=\left(\sum_{i}u_{ik}\right)\left(\sum_ju_{jl}\right)=1$$ The assertion $S_n^{\varepsilon}\subset S_n^{\Gamma_{\varepsilon}}$ follows from the previous lemma.

Noncommutativity of $C(S^{\varepsilon}_n)$
------------------------------------------

Observe that the $C^*$-algebras $C(G^{\varepsilon})$ may collapse to something very small and they might be commutative. This depends on the particular choice of the matrix ${\varepsilon}$ as may 