# Is there a way to classify all metabelian finite groups $G$, such that $\operatorname{Aut}(G) \cong G$?

Is there a way to classify all metabelian finite groups $$G$$, such that $$\operatorname{Aut}(G) \cong G$$?

I know that the trivial group is the only abelian group that satisfies this condition. I also know two non-abelian groups that satisfy this condition: $$S_3$$ and $$D_4$$. But I do not know if there are any other groups.

Any help will be appreciated.

EDIT: Now I also know that $$Hol(Z_n)$$ satisfies this condition for every odd natural $$n$$. But still, is there anything else?

• There are others, see here, and its linked MO-questions, but I do not know a classification. – Dietrich Burde Jul 23 '18 at 13:20

Actually it is possible to classify all finite metabelian complete groups. It was done by T.M.Gagen and D.J.S. Robinson in the article "Finite metabelian groups with no outer automorphism". There they show, that a finite metabelian group is complete iff it is a direct product of holomorphs of cyclic groups of different odd primary order.

That is the strongest result, available to us now, as the answer to the question "Do there exist finite non-complete groups $$G$$, such that $$Aut(G) \cong G$$ and $$G$$ is not isomorphic to $$D_4$$?" is currently unknown.