TITLE: Example of an outer automorphism that maps all elements to conjugates.
QUESTION [8 upvotes]: While answering this question my mind wandered a bit, and it got me wondering about the following question:

Does there exist a (finite) group $G$ with an outer automorphism $\varphi$ such that $g$ and $\varphi(g)$ are conjugate for all $g\in G$?

The word 'finite' is in parentheses because I'd be interested in any example of such a group, but I'd prefer an example with a finite group. I'm convinced such an example must exists, as otherwise this would give a seemingly very weak sufficient condition for an automorphism to be inner, which I would find hard to believe. But I don't mind being surprised.
I've checked a few very non-abelian groups and a few small groups, but I haven't been able to find any such outer automorphisms. If you know of an example, or have some good ideas of where to look for one, I'd be glad to hear.

REPLY [4 votes]: The infinite alternating group $A_\infty$ of even permutations on a countable set (fixing all but finitely many elements) has an outer automorphism of conjugation by some odd permutation $g$, but for any fixed $x\in A_\infty$ we can find some transposition $t$ in the set of fixed points of both $x$ and $g$, and then $tgxg^{-1}t^{-1}=gxg^{-1}$, so $x$ is conjugate (via the even permutation $tg$) to $gxg^{-1}$, even though $g$ is not in $A_\infty$.