divisibility of the order of Aut(G) of a p-group If you were given a p-group G of order $p^{n}$, and you were interested in finding the possible divisors of |Aut(G)|, how would one go about doing this? For instance, could you prove that $p^{n}-1$ divides |Aut(G)|? And finally what if G was not necessarily a p-group, how would one generalize the divisibility of the order of its automorphism group?
 A: Tthe question is probably too broad, but I can answer the part of it referring to $p$-groups $G$ with $|{\rm Aut}(G)|$ divisible by $p^n-1$. I thought at first that only the elementary abelian $p$-group had that property, but interestingly there are just two other examples, namely $C_4 \times C_2^4$, and $Q_8 \times C_2^3$, both of order $64$.
By a result of Zsigmondy, if $p$ is prime and $n>1$, then there is usually a prime $q$ dividing $p^n-1$ that does not divide $p^k-1$ for any $k<n$. The exceptions are when $n=2$ and $p$ is a Fermat prime, and when $p=2$, $n=6$.
Suppose that the $p$-group $G$ has $|{\rm Aut}(G)|$ divisible by $p^n-1$, and suppose that $G$ is not elementary abelian. Then its Frattini subgroup $\Phi(G)$ is nontrivial, and $G/\Phi(G)$ is elementary abelian of order $p^k$ with $k<n$.
If there is a prime $q$ as in Zsigmondy's result, then $q$ does not divide $|{\rm Aut}(G/\Phi(G))| = |{\rm GL}(k,p)|$. So an element of ${\rm Aut}(G)$ of order $q$ would act trivially on $G/\Phi(G)$. But then by a result of Burnside, this automorphism would be trivial, contradiction.
In the exceptional cases when Zsigmondy's result does not hold, when $n=2$, the only possible $G$ is the cyclic group, which has order $p(p-1)$, not divisible by $p^2-1$.
This leaves the case $|G|=2^6$. Since $|{\rm GL}(3,2)|$ has no subgroups of order $9$, we must have $k \ge 4$, so $|\Phi(G)|\le 4$. I was lazy here and finished it off with a computer calculation, although the result could probably be proved by hand.
The only groups of order $64$ with $|\Phi(G)| \le 4$ and  automorphism groups with order divisible by $63$ are $\mathtt{SmallGroup}(64,260) \cong C_4 \times C_2^4$ and $\mathtt{SmallGroup}(64,262) \cong Q_8 \times C_2^3$. Note that the second of these has a subgroup of order $63$, but the first does not.
A: The famous American group theoretician George Abraham Miller constructed in 1913 a non-abelian group of order $2^6$, whose automorphism group is elementary abelian of order $2^7$ (published as Automorphisms  of  Period  Two of  an Abelian Group Whose  Order Is a Power of Two, Tôhoku  J.,  10:118-127).
