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In Scott's classic textbook on Group Theory, he asks:

Suppose that $G$ is a finite group. Is the sequence of isomorphism types of the groups $Aut^{(n)}(G)$ for $n \in \mathbb{N}$ eventually periodic?

Here $Aut^{(2)}(G) = Aut(Aut(G))$ etc. Equivalently, is the sequence $|Aut^{(n)}(G)|$ always bounded above?

It apparently remains opens whether the sequence of automorphism types of $Aut^{(n)}(G)$ is in fact always eventually constant. (A wonderful theorem of Wielandt says that if $G$ is a finite centerless group, then the sequence is eventually constant.) So I would like to ask:

Does there exists a finite group such that $Aut(G) \not \cong G$ but $Aut^{(n)}(G) \cong G$ for some $n \geq 2$?

Edit: Joel has pointed out that my question is even open for infinite groups. This sounds like an interesting question which doesn't seem amenable to the standard tricks.

In Scott's classic textbook on Group Theory, he asks:

Suppose that $G$ is a finite group. Is the sequence of isomorphism types of the groups $Aut^{(n)}(G)$ for $n \in \mathbb{N}$ eventually periodic?

Here $Aut^{(2)}(G) = Aut(Aut(G))$ etc. Equivalently, is the sequence $|Aut^{(n)}(G)|$ always bounded above?

It apparently remains opens whether the sequence of automorphism types of $Aut^{(n)}(G)$ is in fact always eventually constant. (A wonderful theorem of Wielandt says that if $G$ is a finite centerless group, then the sequence is eventually constant.) So I would like to ask:

Does there exists a finite group such that $Aut(G) \not \cong G$ but $Aut^{(n)}(G) \cong G$ for some $n \geq 2$?

Edit: Joel has pointed out that my question is perhaps even open for infinite groups. This sounds like an interesting question which doesn't seem amenable to the standard tricks.