Bijection on conjugate equivalence induce isomorphism? Given group $G$, we denote $G'=G/$~, where "~" is conjugate equivalence ($a$~$b$ iff there exists $c$, s.t. $a=c^{-1}bc$) . Then, a group homomorphism $f:G\to H$ induces a map in  sets of conjugate equivalent classes $f':G'\to H'$.
Now, suppose the induced map $f'$ is a bijection, is it true that the original group homomorphism $f$ is an isomorphism?
I think the hard part is proving $f$ is surjective. And in finite case, I tried to use Class Equation of Group to work it out. But I still got stuck and couldn't go any further. Hope someone could help. Thanks!
 A: If $H$ is a finite group, the answer is YES (I don't know how to conclude if $H$ is infinite).
If $g\in G$, let use denote by $\gamma_G(g)$ its equivalence class, i.e. the set $\gamma_G(g)=\{ c^{-1}g c\mid c\in G\}$, and let's use a similar notation for $H$.
Let $f:G\to H$ be a group morphism such that the
 map $f':\gamma_G(g)\in G'\mapsto \gamma_H(f(g))\in H'$  is bijective.
First step. $f$ is injective (it works even if $H$ is not supposed to be finite).
For, let $g\in G$ such that $f(g)=1_H$. Then $f'(\gamma_G(g))=\gamma_H(f(g))=\gamma_H(1_H)=f'(\gamma_G(1_G))$, and by assumption, $\gamma_G(g)=\gamma_G(1_G))=\{1_G\}$. Thus , $g$ is conjugate to $1_G$, which easily implies that $g=1_G$.
Second step. Let $K=im(f)$. Then $H=\displaystyle\bigcup_{h\in H}hKh^{-1}$.
Indeed, let $h\in H$. By assumption, there exists $g\in G$ such that $\gamma_H(h)=f'(\gamma_G(g))=\gamma_H(f(g))$. It means that $h$ is conjugate to $f(g)\in K$, hence the desired equality.
Third step. Assume that $H$ is finite (and thus so is $G$, since $f$ is injective). Then $K=H$ , that is , $f$ is surjective.
Indeed, it is a classical fact that a finite group cannot be the union of conjugates of a proper subgroup.
Let's give a quick proof of this fact. Let $U= \displaystyle\bigcup_{h\in H}hKh^{-1}$.
Let $h_1,\ldots,h_r$ be the representatives of the left cosets of $H$ modulo $K$. Then it is easy to see that $U\setminus\{1_H\}=\displaystyle\bigcup_{i=1}^r (h_iKh_i^{-1}\setminus\{1_H\})$.
In particular, $\vert U\vert-1\leq r(\vert K\vert -1)=\vert H\vert-[H:K]$. If $U=H$, then $[H:K]\leq 1$, so $[H:K]=1$ and $K=H$.  
Note that $GL_2(\mathbb{C})$ is the union of the conjugates of the subgroup of upper triangular matrices, so the third step does not hold in general in the infinite case.
