TITLE: Does $\mathrm{Aut}(\mathrm{Aut}(...\mathrm{Aut}(G)...))$ stabilize?
QUESTION [163 upvotes]: Purely for fun, I was playing around with iteratively applying $\DeclareMathOperator{\Aut}{Aut}\Aut$ to a group $G$; that is, studying groups of the form
$$ {\Aut}^n(G):= \Aut(\Aut(...\Aut(G)...)) $$
Some quick results:

For finitely-generated abelian groups, it isn't hard to see that this sequence eventually arrives at trivial group.
For $S_n$, $n\neq 2,6$, the group has no center and no outer automorphisms, and so the conjugation action provides an isomorphism $G\simeq \Aut(G)$.
Furthermore, I believe that if $G$ is a non-abelian finite simple group, then $\Aut(\Aut(G))\simeq \Aut(G)$, though this is based on hearsay.
If one considers topological automorphisms of topological groups, then $\Aut(\mathbb{R})\simeq \mathbb{Z}/2\mathbb{Z}\times \mathbb{R}$, and so $$\Aut(\Aut(\mathbb{R}))\simeq \Aut(\mathbb{Z}/2\mathbb{Z}\times \mathbb{R})\simeq\mathbb{Z}/2\mathbb{Z}\times \mathbb{R}.$$

When the sequence $\Aut^n(G)$ is constant for sufficiently large $n$, we will say the sequence stabilizes.  Despite my best efforts, I have been unable to find a group $G$ such that the sequence $\Aut^n(G)$ is provably non-stabilizing.  Is this possible?
A slightly deeper question is whether there are groups $G$ such that the sequence becomes periodic after some amount of time.  That is, $\Aut^n(G)\simeq \Aut^{n+p}(G)$ for some $p$ and for $n$ large enough, but $\Aut^n(G)\not\simeq \Aut^{n+m}(G)$ for $m$ between 0 and $p$.  A simple way to produce such an example would be to give two groups $G \neq H$ such that $\Aut(G)\simeq H$ and $\Aut(H)\simeq G$.  Does anyone know an example of such a pair?

REPLY [14 votes]: It seems that $Aut(G)$ is complete for any simple (not necessarily finite) group. The proof of this is in Rotman's book, but it is very short.