$k\geq 1$ such that $H_{m}=H_{m+1}$ for any $m\geq k$. As $Z_{G}(H^{0})=\cap_{m\geq 1}Z_{G}(H^{0}[m])$, we get $Z_{G}(H^{0})=Z_{G}(H^{0}[m])$ for any $m\geq k$.

There is an exact sequence $1\rightarrow H^{0}\rightarrow H\rightarrow H/H^{0}\rightarrow 1.$ Then, the group $H$ is determined by a homomorphism $\phi: H/H^{0}\rightarrow\Aut(H^{0})$ and a class $\alpha\in
H^{2}_{\phi}(H/H^{0},H^{0})$. Write $n=|H/H^{0}|$. Then, $n\alpha=0$. There is an exact sequence $$1\rightarrow
H^{0}[n]\rightarrow H^{0}\rightarrow H^{0}\rightarrow 1,$$ where the homomorphism $H^{0}\rightarrow H^{0}$ is $$\psi_{n}(x)=x^{n}, \forall x\in H^{0}.$$ Due to $\psi_{n}(\alpha)=n\alpha=0$, the class $\alpha$ comes from $H^{2}_{\phi}(H/H^{0},H^{0}[n])$. That means there is a finite subgroup $J'$ of $H$ with $J'\cap H^{0}=H^{0}[n]$ and $H=H^{0}J'$. Set $J=J'H^{0}[nk]$. For any $m\geq 1$, set $J_{m}=JH^{0}[nkm]$. Since $G$ is acceptable, there exists $g\in G$ such that $$\phi_{2}|_{J}=\Ad(g)\circ\phi_{1}|_{J}.$$ Set $\phi'_{2}=\Ad(g^{-1})\circ\phi_2$. Then, $\phi'_{2}|_{J}=\phi_{1}|_{J}$. For any $m\geq 1$, by the acceptableness of $G$ there exists $g_{m}\in G$ such that $$\phi_2'|_{J_{m}}=\Ad(g_{m})\circ\phi_{1}|_{J_{m}}.$$ Then, $$g_{m}\in Z_{G}(\phi_{1}(H^{0}[nk]))=Z_{G}(H^{0}).$$ Thus, $\phi'_{2}|_{H^{0}}=\phi_{1}|_{H^{0}}$. By $H=H^{0}J$, we get $\phi'_{2}=\phi_{1}$. Hence, $\phi_1$ and $\phi_2$ are globally conjugate.

Further reduction
-----------------

The following lemma is trivial.

\[L:product\] Let $G_1$ and $G_2$ are compact Lie groups, then $G_1\times G_2$ is acceptable (or strongly acceptable) if and only if both of $G_1$ and $G_2$ are.

The following statement follows from [@Larsen1 Prop. 1.4].

\[L:Larsen\] Let $G$ be a connected compact Lie 