Given a finite cyclic group $G$, we denote by $L(G)$ the lattice of its subgroups, and by $\mathop{\rm Aut}(G)$ the automorphism group of $G$. Let $H$ be any group. Assume that $L(G)\cong L(H)$ and $\mathop{\rm Aut}(G)\cong\mathop{\rm Aut}(H)$. Is it true that $G\cong H$?
1 Answer
$\def\ZZ{\mathbb Z}$If $G\cong \ZZ/n\ZZ$, then the lattice $L(G)$ tells the number of prime factors of $n$ together their exponents, while $\mathop{\rm Aut} G\cong (\ZZ/n\ZZ)^*$, which can be decomposed into a direct product using the Chinese remainders theorem.
Now it is easy to construct a counterexample. Since $(\ZZ/27\ZZ)^*\cong(\ZZ/19\ZZ)^*\cong\ZZ/18\ZZ$ and $(\ZZ/125\ZZ)^*\cong(\ZZ/101\ZZ)^*\cong\ZZ/100\ZZ$, we have $(\ZZ/(27\cdot 101\ZZ))^*\cong(\ZZ/(19\cdot 125\ZZ))^*$ and also $L(\ZZ/(27\cdot 101\ZZ))\cong L(\ZZ/(19\cdot 125\ZZ))$.

$\begingroup$ Sorry, the previous version used $(\ZZ/8\ZZ)^*$ which is not cyclic... $\endgroup$ Commented Dec 22, 2015 at 10:27