TITLE: Is an isomorphism sends any element to one of its conjugation elements inner?
QUESTION [7 upvotes]: If $G$ is a finite group, $f$ is an isomorphism from $G$ to $G$, s.t for any $x \in G$, there exists $y \in G$, and $f(x)=yxy^{-1}$ (note the order of quantifiers), is $f$ an inner isomorphism?

REPLY [5 votes]: Such automorphisms also are known under the name Almost-inner automorphisms. This concept goes back to Burnside for finite groups, and then later was studied in the question "Can you hear the shape of the drum" for isospectral but not isometric manifolds $G/\Gamma$, with a solvable or nilpotent Lie group $G$ with discrete cocompact subgroup $\Gamma$, and almost inner automorphisms of $G$, which are not inner. For a survey see the introduction in our paper Almost-inner derivations of Lie algebras.
In the finite group case, the above thesis of Potter gives examples of groups admitting non-inner almost inner automorphisms, e.g., Burnside's group. In the last chapter, Potter also describes how to obtain a lower bound on the number of isospectral nonisometric Riemann surfaces.