Is there a way to describe all finite groups $G$, such that $Aut(G) = D_4$?

Two groups, that definitely satisfy that condition are $D_4$ itself and $\mathbb{C}_2 \times \mathbb{C}_4$.

I have read somewhere, that those two groups are the only two groups, that satisfy that condition. There was no proof of this statement given, however, so I do not know whether it is true or false (And if it is true, it would be interesting to know the proof).

Any help will be appreciated.

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    $\begingroup$ You asked a similar question about $S_3$ last month at math.stackexchange.com/questions/2875892. Did you try using the ideas in the answers there? If so, tell us what you found. If not, try to see what those ideas can do for you in this new setting. $\endgroup$ – KCd Sep 23 '18 at 14:23

Yes, this is true. This is Theorem 6.3(d) in the paper, “On solving the equation $\operatorname{Aut}(X) = G$“, by H. K. Iyer, Rocky Mountain J. Math. 9, No. 4, (1979), 653-670. The theorem lists all finite groups with automorphism group a dihedral group. There are many other results of a similar nature in the paper as well.


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