I was asked the following question:
Give an example of two non-isomorphic $\mathbb Z[i]$-modules which are isomorphic as $\mathbb Z$-modules (abelian groups).
There are several similar questions on math.stackexchange, but I was not able to adapt answers for my problem.
I need a hint to start. Thanks!
So the question is actually: find an abelian group $G$ and two automorphisms $\phi$,$\psi$ with square $-id$ such that no automorphism of $G$ maps $\phi$ to $\psi$.
In other words, we want to find an abelian group $G$ and $\phi,\psi \in Aut(G)$ that are not in the same conjugacy class, with $\phi^2=\psi^2=-1$.
We are done as soon as $Aut(G)$ is abelian and we can choose $\phi \neq \psi$.
For instance, we can take $\mathbb{Z}/5\mathbb{Z}$ as its automorphism group is isomorphic to $\mathbb{Z}/4\mathbb{Z}$ so is abelian, and take $\phi=2id$, $\psi=3id$.
In terms of the original problem: define $M_2$ (resp. $M_3$) to be the $\mathbb{Z}[i]$ module $\mathbb{F}_5$ where $i$ acts by multiplication by $2$ (resp. $3$). So $M_2$ and $M_3$ are isomorphic over $\mathbb{Z}$, but they are not isomorphic over $\mathbb{Z}[i]$. Indeed, the ideal $I$ of elements acting as $0$ on $M_k$ is $(5,i-k)$, and $(2-i)=(5,i-2) \neq (5,i-3)=(5,2+i)=(2+i)$.