Iterated automorphism groups of finite groups

Let $\mathcal{G}$ be the set of isomorphism classes of finite groups.

There is an operation $\mathrm{Aut} : \mathcal{G} \rightarrow \mathcal{G}$ which gives the automorphism group of a given group, up to isomorphism. In the most general setting, my question is:

What possible eventual behaviour can arise from iterating the operation $\mathrm{Aut}$?

More precisely, there is a trichotomy of possible eventual behaviours:

• Static: a group such as $D_8$ or $S_8$, which is isomorphic to its automorphism group.
• Periodic: a group which enters a cycle of length $\geq 2$.
• Divergent: a group which never enters a cycle, and therefore grows without bound.

I can exhibit lots of examples of static groups, and groups which eventually evolve into static groups after many steps, such as:

$$C_{2879} \mapsto C_{2878} \mapsto C_{1438} \mapsto C_{718} \mapsto C_{358} \mapsto C_{178} \mapsto C_{88} \\ \mapsto C_2 \times C_2 \times C_2 \times C_5 \mapsto PSL(2,7) \times C_4 \mapsto PGL(2, 7) \times C_2$$

However, I don't have any examples of periodic or divergent groups.

I conjecture that $C_{41}$ does diverge, simply because after several iterations it spawns a direct sum involving lots of copies of $C_2$:

$$C_{41} \mapsto C_{40} \mapsto C_4 \times C_2 \times C_2 \mapsto D_4 \mapsto F_4/Z \times C_2 \\ \mapsto (S_4 \wr C_2) \times C_2 \\ \mapsto (S_4 \wr C_2) \times C_2 \times C_2 \\ \mapsto (S_4 \wr C_2) \times S_4 \times C_2 \times C_2 \\ \mapsto (S_4 \wr C_2) \times S_4 \times S_4 \times C_2 \times C_2 \times C_2 \times C_2$$

Finally, for the most ambitious question of all:

• Does there exist an algorithm which, when given a description of a Turing machine, outputs a finite group which is divergent if and only if the Turing machine does not halt?
• For your last question, in what form do you want the finite group to be given? If only as a presentation, then one can do this just by considering that the question of whether a given presentation is the trivial group or not is Turing equivalent to the halting problem. – Joel David Hamkins Mar 2 '18 at 13:09
• Another relevant question: math.stackexchange.com/questions/1096939/… – Nick Gill Mar 2 '18 at 14:03
• @JoelDavidHamkins It's finite, so it may as well be given as a Cayley table. – Adam P. Goucher Mar 2 '18 at 18:03