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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 $\text{Aut}(G)$.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$C_G(N)$C_A(N)$ is the kernel of the restriction of $G$ A$to$N$, all what remain to show is that$C_G(N)$C_A(N)$ 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 (non-canonically) isomorphic to $G/K$.

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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 $\text{Aut}(G)$.

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 $C_G(N)$ is the kernel of the restriction of $G$ to $N$, all what remain to show is that $C_G(N)$ is solvable too.