TITLE: On $p$-groups with abelian automorphism group
QUESTION [8 upvotes]: Let $G$ be a $p$-group of order $p^{n}\geq p^{7}$ and its automorphism group is elementary abelian $p$-group. Then, it is clear that $G$ is nilpotent of class $2$. However, the converse is not true in general (take for example the unitriangular group of $3\times 3$ upper triangular matrices over $\mathbb{F}_{p}$). In fact, Marta Morigi proves that there is a $p$-group of order $p^{7}$ and class $2$ and its automorphism group is elementary abelian $p$-subgroup of order $p^{12}$. Furthermore, I think the converse can be true for 2-groups $G$ with abelian direct factor but I don't know how to do this.
Are there any assumption to add to p-groups of class $2$ to get the converse for arbitrarily integer $n$?
Let $p$ be an odd prime and $G$ be a purely non-abelian p-group of class $2$ and order $p^{n}\geq p^{7}$. Does the automorphism group of $G$ is abelian?
Are there some other class of $p$-groups with elementary abelian automorphism $p$-groups?.
Any response or reference may be very helpful. Thank you in advance.

REPLY [6 votes]: This is just an amplification of my comment above.
Theorem 3.3 of this this survey article, A Survey on Automorphism Groups of Finite p-Groups, by Geir T. Helleloid (2006) describes a result published in
U. M. Webb, The number of stem covers of an elementary abelian p-group,
Math. Z. 182 (1983), no. 3, 327–337.
A stem cover of a group $Q$ is a maximal stem extension of $Q$; that is, a maximal  (under group extensions) group $G$ with normal subgroup $N$ such that $G/N \cong Q$ and $N \le Z(G) \cap [G,G]$. (So, in a stem cover, $N$ is isomorphic to the Schur Multiplier of $Q$.)
This result says that, for $p$ an odd prime, the proportion of stem covers of an elementary abelian group of order $p^n$ for which the automorphism group is an elementary abelian $p$-group approaches 1 as $n \to \infty$. Elementary abelian gropups have lots of distinct stem covers, so this is producing plenty of examples of the type you are looking for.