conjugacy classes in $G$. We know that $G^{\#}$ has a topology of a compact space (actually it is the union of finitely many compact orbifolds), and the natural map $\pi: G
\rightarrow G^{\#}$ is continuous. For an element $g\in G$, write $[g]=\pi(g)\in G^{\#}$ for its conjugacy class. The map $\pi$ has a further property: for any sequence $\{g_{n}\}_{n\geq 1}\subset G$, $$\label{Eq1}\lim_{n\to\infty}g_{n}=1\Leftrightarrow\lim_{n\to\infty}[g_{n}]=[1].$$

Let $\Gamma$ be a compact group and $\phi_1,\phi_2:\Gamma\rightarrow G$ be two element-conjugate homomorphisms. Then for any $x\in\Gamma$, $\phi_{1}(x)=1$ if and only if $\phi_{2}(x)=1$. Thus, $\ker\phi_1=\ker\phi_2$. Considering $\Gamma/\ker\phi_{1}$ instead, we may assume that $\phi_1,\phi_2$ are injections. Write $H=
\overline{\phi_{1}(\Gamma)}$ for the closure of the image of $\phi_1$, which is a closed subgroup of $G$. We also use $\phi_1$ to denote the injection $\Gamma\rightarrow H$. We first show that there is another injection $\psi': H\rightarrow G$ which is element-conjugate to the inclusion $\psi: H\hookrightarrow G$, such that $\phi_2=\psi'\circ\phi_1$. For any $h\in H$, choose a sequence $\{\gamma_{n}\}_{n\geq 1}\subset
\Gamma$ such that $$\lim_{n\to\infty}\phi_{1}(\gamma_{n})=h.$$ Then, $$\lim_{n,m\to\infty}\phi_{1}
(\gamma_{n}^{-1}\gamma_{m})=1.$$ By Property (\[Eq1\]), we get $$\lim_{n,m\to\infty}[\phi_{1}
(\gamma_{n}^{-1}\gamma_{m})]=1.$$ As $\phi_1$ and $\phi_2$ are element-conjugate, we get $$\lim_{n,m\to
\infty}[\phi_{2}(\gamma_{n}^{-1}\gamma_{m})]=1.$$ By Property (\[Eq1\]) again, we get $$\lim_{n,m\to
\infty}\phi_{2}(\gamma_{n}^{-1}\gamma_{m})=1.$$ Thus, $\{\phi_{2}(\gamma_{n})\}_{n\geq 1}$ is a Cauchy sequence in $G$. Define $$\psi'(h)=\lim_{n\to\infty}\phi_{2}(\gamma_{n}).$$ Using 