TITLE: $a,b$ are conjugate if every $f(a),f(b)$ are
QUESTION [9 upvotes]: Let $G$ be a finite group, let $a,b\in G$. For every homomorphism $f$ from $G$ to some symmetric group $S_n$, we are given that $f(a)$ and $f(b)$ are conjugate in $S_n$. Prove or give a counterexample to disprove that $a,b$ are conjugate in $G$.

Here is what i have thought so far: We know that an even permutation $\sigma$'s conjugate class in $S_n$ are consist of even permutations. And if we consider the conjugate class of $\sigma$ in alternating group $A_n$, one of two things will happen, either conjugate class doesn't change or it splits into two conjugate classes. For example there exist two 5-cycles $a,b$ in $A_5$ don't conjugate to each other. But they are conjugate in $S_5$. So do we have for every $f:A_5\to S_n$, $f(a)$ and $f(b)$ are conjugate?

REPLY [4 votes]: In $S_n$, the conjugacy class of an element is determined by its cycle type, so every element is conjugate to its inverse. We can get a counterexample from any group in which there is an element not conjugate to its inverse, for instance $G=\mathbb{Z}/3\mathbb{Z}$, $a=1$, $b=2$.