I have the two following groups
$G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^5$, where $A=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&0&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$ and
$G_B=\mathbb{Z}\ltimes_B \mathbb{Z}^5$, where $B=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&1&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$.
The product is given (for example in $G_A$) by $(k,m)\cdot(\ell,n)=(k+\ell, m+A^k n)$.
Problem: Decide if $G_A$ is isomorphic to $G_B$ or not.
My thoughts: I think strongly that they are not isomorphic. The matrices $A$ and $B$ are both of order 4, they're not conjugate in $\mathsf{GL}(n,\mathbb{Z})$ (neither $B$ and $A^{-1}$) but they are conjugate in $\mathsf{GL}(n,\mathbb{Q})$. In some other cases, I've seen that they're not isomorphic by computing the abelianization, but in this case both have the same abelianization, namely $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$. Even worse, both have 1 as an eigenvalue.
In my previous MO question there is a counterexample for the implication "$G_A\cong G_B\Rightarrow A\sim B^{\pm 1}$" so I cannot use that.
Can anyone give me any other ideas to prove that they aren't isomorphic? Or maybe to prove that they are isomorphic (if they were). Now I posted it in this MO question
Thanks!