Let $G$ be an abelian (not elementary) finite $p$group. In what conditions the automorphism group of $G$ is solvable?

I believe that for $p\ge5$, these are the direct products of cyclic groups of pairwise distinct orders, for if $C_{p^e}\times C_{p^e}$ is a direct factor of $G$, then $GL(2,p)$ is a homomorphic image of a subgroup of $A=\text{Aut}(G)$. Similarly for $p=2$ or $3$, I expect the groups $G$ you are looking for are those where no order in the direct product appears more than $2$ times. In order to show that these groups indeed have a solvable automorphism group, you probably may apply induction: Let $N$ be the subgroup of $G$ generated by the $p$th powers in $G$. Then $N$ has the same shape as above (with the order of each direct factor divided by $p$). So $\text{Aut}(N)$ is solvable by induction. As $C_A(N)$ is the kernel of the restriction of $A$ to $N$, all what remain to show is that $C_A(N)$ is solvable too. Alternatively, one could try to use the induction hypothesis for $G/K$, where $K$ is generated by the elements of order $p$. Note that $N$ is (noncanonically) isomorphic to $G/K$. 


The Frattini quotient G/Phi(G) is the maximal elementary abelian quotient of G; the group of automorphisms of G which act trivially on G/Phi(G) is a pgroup, and so Aut(G) is solvable if and only if its image in the linear group Aut(G/Phi(G)) is; that's at least a partial answer. Also, this holds for general prop groups, not only for finite abelian pgroups. 

