Let $G$ be a finite group whose center $Z(G)$ is trivial. Suppose that the group $\text{Out}(G)$ of outer automorphisms is nontrivial.
Question: Does there always exist an $f \in \text{Aut}(G)$ and a $g \in G$ such that $g$ and $f(g)$ do not lie in the same conjugacy class?
This question can be reformulated in the following way. The automorphism group $\text{Aut}(G)$ acts on the set of conjugacy classes of $G$. As the inner automorphism group $\text{Inn}(G)$ acts trivially on this set, the action descends to the quotient $\text{Out}(G)$. With the same assumptions as above, the question then becomes:
Question: Is the action of $\text{Out}(G)$ on the set of conjugacy classes always nontrivial?
Remark: I restrict myself to the case that $G$ is centerless because this case is of main importance to me. I am not sure though whether there is an easy answer or counterexample to the question in the case that we do not assume $G$ to be centerless.
What I've got so far: the question is positive for the symmetric group $S_{n}$ on $n$ letters for all $n \geq 3$. Indeed, only for $n = 6$ there exists an outer automorphism and this automorphism interchanges the conjugacy class of a cycle of cycle type $(ab)$ and the conjugacy class of a cycle of cycle type $(cd)(ef)(gh)$. Here, $a \neq b$ and the $c, d, e, f, g$ and $h$ are distinct.
This question is first asked by Burnside;
Such automorphisms are called as class preserving automorphisms and denoted by $A_c$ or $Aut_c(G)$. As you also notice $Inn(G) \leq A_c\leq Aut(G)$. For some groups two of such groups can be equal. Check this link. If you search as "Class preserving automorphisms", you can find many papers related the this topic.
Edit:
In this paper,