Given a finite cyclic group $G$ with order $n=p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}$, where $p_i$s are distinct prime numbers $n_i>1$ for all $i$. Let $H$ be any abelian group. Assume that Aut$(G)≅$Aut$(H)$. Is it true that $G≅H$?

$\begingroup$ No. Let $G$ have order $3$, and let $H$ be an infinite cyclic group. $\endgroup$– Derek HoltCommented Dec 23, 2015 at 14:35

$\begingroup$ Another example: the cyclic groups of order $15$ and $16$ have isomorphic automorphism groups. $\endgroup$– Derek HoltCommented Dec 23, 2015 at 14:38

1$\begingroup$ I am voting to close on the grounds that there are very easy counterexamples. $\endgroup$– Derek HoltCommented Dec 23, 2015 at 14:41
1 Answer
No. For example, if $G$ is cyclic of order $900 = 2^{2} \cdot 3^{2} \cdot 5^{2}$, and $H$ is cyclic of order $496 = 2^{4} \cdot 31$, then ${\rm Aut}(G) \cong {\rm Aut}(H)$.
It's not too hard to see that if $H$ is abelian, and ${\rm Aut}(H)$ is abelian, then $H$ must be cyclic. So it follows that if $G$ is cyclic, $H$ is abelian, and ${\rm Aut}(G) \cong {\rm Aut}(H)$, then $H$ is cyclic.
Finally, if you assume that $G$ and $H$ are both cyclic and have powerful order and ${\rm Aut}(G) = {\rm Aut}(H)$, then $G \cong H$. This is because if $p$ is the largest prime factor of ${\rm Aut}(G)$ then ${\rm ord}_{p}({\rm Aut}(G)) = {\rm ord}_{p}(G)  1$. It follows then that the powers of $p$ dividing $G$ and $H$ are the same, and one obtains the result by induction.