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This is problem 11.123 in the Kourovka notebook:

For a given group $G$, define the following sequence of groups: $A_1(G) = G$, $A_{i+1}(G) = \operatorname{Aut}(A_i(G))$. Does there exist a finite group $G$ for which this sequence contains infinitely many non-isomorphic groups? (M. Short)

I'm familiar with Wielandt's negative answer in the special case that $\mathbf{Z}(G)=1$. Have any other advances been made?

LSpice
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