I wondered whether it is possible to find two finitely generated, virtually Abelian, torsionfree groups $G,H$ that are not isomorphic but that become isomorphic after crossing with $\mathbb{Z}$. I have the following candidates:

Consider $K:=(\mathbb{Z}[t]/(t^5+1))\rtimes_{\cdot t} \mathbb{Z}$. Let $\varphi$ be the automorphism of $K$ given by $\cdot t$ on $\mathbb{Z}[t]/(t^5+1)$ and $(0,s)\mapsto (1,s)$ , where $s$ denotes a generator of the other copy of $\mathbb{Z}$. My candidates are $G:=K\rtimes\mathbb{Z}$ and $H:=K\rtimes_{\varphi^3}\mathbb{Z}$.

*Are these two groups isomorphic ?*

Each one contains a finite index subgroup isomorphic to the other one. Crossing with $\mathbb{Z}$ gives $K\rtimes \mathbb{Z^2}$ where a basis of $\mathbb{Z}^2$ acts by $\varphi,\mbox{id}$ respectively $\varphi^3,\mbox{id}$. The isomorphism is given by a base change of $\mathbb{Z}^2$.

$G$ contains a finite index subgroup isomorphic to $H$ and vice versa. If it turns out that they are actually isomorphic, one might still hope to get an example by replacing $5$ with a bigger number (that is coprime to 3 to make the base change work).

Related: Hirshon, some cancellation theorems with applications to nilpotent groups (The example given there is torsionfree, nilpotent, but maybe not virtually Abelian).