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?
