Isomorphism of $\mathbb{Z}\ltimes_A \mathbb{Z}^m$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^m$

Question

Let $A$ and $B$ be matrices of finite order with integer coefficients.

Let $n\in\mathbb{N}$ and let $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^n$ be the semidirect product, where the action is $\varphi(n)\cdot (m_1,\ldots,m_n)=A^n (m_1,\ldots,m_n)$, and similarly with $B$.

It is easy to construct an isomorphism between $G_A$ and $G_B$ if $A$ is conjugate in $GL(n,\mathbb{Z})$ to $B$ or $B^{-1}$.

But, this is also a necessary condition? I mean, does $G_A\cong G_B$ implies $A\cong B$ or $A\cong B^{-1}$ in $GL(n,\mathbb{Z})$ or is there a counterexample?

I've seen in A necessary condition for two semi-direct products to be isomorphic. that it is true if A and B are hyperbolic, i.e none of their eigenvalues have module 1, but it isn't the case.

Thank you!

Note that your assumption implies that $A,B$ have finite order. — YCor, Jan 08 '20 at 17:18
I think it would fit on MO (I'd recommend mention explicitly that $A,B$ have finite order). Also a remark, denoting by $G_A$ this group, is that $G_A$ is virtually abelian iff $A$ has finite order. In particular, if $A$ has finite order and $G_A\simeq G_B$ then $B$ also has finite order. — YCor, Jan 09 '20 at 18:51
@YCor thank you for your recommendations, I will do that. By the way, what's the difference between MO and MSE? — Alejandro Tolcachier, Jan 09 '20 at 19:28
https://math.stackexchange.com/help/on-topic, https://mathoverflow.net/help/on-topic — YCor, Jan 09 '20 at 21:04
Now posted on MO: Isomorphism of $\mathbb{Z}\ltimes_A \mathbb{Z}^m$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^m$. (It is useful to mention the link to the other site. The people interested in this topic can find what was said in the other post. And potential answerers will not repeat the stuff which has been already said in the other place. You can also check out general recommendations concerning [meta-tag:cross-posting].) — Martin Sleziak, Jan 10 '20 at 06:02

Isomorphism of $\mathbb{Z}\ltimes_A \mathbb{Z}^m$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^m$

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