TITLE: What does the Outer automorphism group act on? If an automorphism preserves conjugacy classes, is it inner?
QUESTION [5 upvotes]: (Let $x^G:=\{gxg^{-1}:g\in G\}$ denote the conjugacy class of $x$)
Am I right in thinking $\text{Out }G := \text{Aut }G/\text{Inn }G$ acts on the conjugacy classes of G, $\{x^G:x\in G\}$? What else could I think of it acting on (trying to get some intuition)?
In addition to that, if $f\in \text{Aut }G$
$$f\in\text{Inn }G\iff(\forall x\in G)f(x^G) = x^G$$
I feel this must be true, but couldn't find anything on it.
If it is true, can we find a weaker condition on the right hand side to check if $f\in\text{Inn }G$? eg. we only require $f(x^G) = x^G$ for $x$ in a set of generators for this to be true.

REPLY [5 votes]: Here is a simple example where the Grossman property is violated:
Let $X$ be an infinite set. A permutation $\alpha$ of $X$ is said to have finite support  if there exists a finite subset $F\subset X$ such that $\alpha$ fixes every element of $X-F$.
The set $Aut_f(X)$ of permutations of $X$ with finite support clearly form a normal subgroup of the group $Aut(X)$ of all permutations of $X$.
Suppose that $\alpha$ is any permutation of $X$. Then for every permutation $\beta\in Aut_f(X)$, the permutations $\alpha \beta \alpha^{-1}$ and $\beta$ are conjugate not only in $Aut(X)$, but also in $Aut_f(X)$ (I leave it to you as an exercise to prove this). At the same time, for each $\alpha$ with infinite support, the automorphism of $Aut_f(X)$ defined by
$$
\beta \mapsto \alpha \beta \alpha^{-1}
$$
is not inner (as an automorphism of $Aut_f(X)$). This is another easy exercise.