Find finite groups $G\cong Aut(G)$

Question

I become interested in this problem because $G\cong Aut(G)$ suggests a special symmetry in $G$ (This kind of group describes its own symmetry).

From $G/Z(G)\cong Inn(G)$ we know complete group is the answer for the simplest case, though this class of group itself is quite weird.

What about the case when $Z(G)$ or $Out(G)$ are non-trivial? $G\cong D_8$ is a good example for non-complete group satisfy $G\cong Aut(G)$.

Is there any method to find such non-complete $Aut$-invariant $G$? Maybe we can try apply $Aut$ iteratively to some G and look for fixed points (some interesting results for centerless or non-abelian simple groups can be found at this MO post).

And what other interesting properties does this class of group have?

Answer 1

Let's call a group $G$ quasicomplete if $Z(G) \ne 1$ and $G \cong {\rm Aut}(G)$.

If $H$ is a complete group with a unique subgroup of index $2$, then $H \times C_2$ is quasicomplete. I have checked that all quasicomplete groups of small order (order less than $768$ so far) apart from $D_8$ have this form. There are lots of $H$ with this property, such as $S_n$ for $n>6$, or ${\rm PGL}(2,p)$ for odd primes $p \ge 5$, or the semidirect product $C_p \rtimes C_{p-1}$ (with faithful action) for odd primes $p$.

Another infinite class of examples is $H \times D_8$, where $H$ is a complete group with no subgroup of index $2$. Again, there are lots of examples of $H$, including some simple groups, such as ${\rm Sp}(2n,2)$ for $n \ge 3$, and there are also complete groups of odd order.

So the question is, are there are any (finite) quasicomplete groups that are not of one of these two types? Of course, this is a more general question than the one about $p$-groups in the Kourovka Notebook.