My problem: with $GAP$ prove there exists a group $G$, $|G|=32$ and maximal subgroup $M$ such that $\text{Aut}(G) \cong \text{Aut}(M)$.
It is clear that $G$ is $p$-group, ($32=2^5$), but with $GAP$ we have:
There are $51$ groups of order $32$.
They are sorted by their ranks.
$1$ is cyclic.
$2\ldots20$ have rank $2$.
$21\ldots44$ have rank $3$.
$45\ldots50$ have rank $4$.
$51$ is elementary Abelian.
Here is list of all Groups of order $32$, but there are $51$ groups of order $32$. My question: which of this group is exactly the answer? How can we find it? How can we find "IdSmallGroup($32$)" with the above property? What is StructureDescription of $G$ and $M$?
I think there is exactly one group. Is this group with this property unique? What is the GAP ID of this group?