Finding isomorphism between $\mathbb{Z}^2\ltimes_{A,B} \mathbb{Z}^4$ and $\mathbb{Z}^2\ltimes_{A,C} \mathbb{Z}^4$ Let $A,B(a,b,c,d)\in\mathsf{GL}(4,\mathbb{Z})$ be given by $$A=\begin{pmatrix} I_2 & \begin{pmatrix} 0&0\\0&1 \end{pmatrix} \\0& -I_2\end{pmatrix},\quad B(a,b,c,d)=\begin{pmatrix} -2a-b & 2c& 0& c\\ -\frac{1+(2a+b)^2}{2c}&2a+b&d&a\\0&0&-b& \frac{1+b^2}{2d}\\ 0&0&-2d&b\end{pmatrix}$$ (which implies $a,b,c,d\in \mathbb{Z}$, $b$ odd, $2c\mid 1+(2a+b)^2$ and $2d\mid 1+b^2$). $A$ has order 2 and $B^2(a,b,c,d)=-Id$ so has order 4.
In general, if we have $A,B$ we can define the group $G_{A,B}$ by $G_{A,B}=\mathbb{Z}^2\ltimes \mathbb{Z}^4$ where the product is given by $(k,\ell,v)\cdot (k',\ell',v')=(k+k',\ell+\ell',v+A^k B^{\ell} v')$.
Problem: Decide if $G_{A,B(a,b,c,d)}$ is isomorphic to $G_{A,B(e,f,g,h)}$ or not.
Thoughts: Actually, I think that the groups are isomorphic for every possible choice of parameters (I've computed some computational invariants (Abelianization, #LowIndexSubgroups, #AutomorphismGroup(PQuotient)) of the groups given by some specific choices of the parameters and they coincide) but I can't find an isomorphism. I have proved the following (for any $A,B\in\mathsf{GL}(n,\mathbb{Z}$)):

*

*$G_{A,B}$ is isomorphic to $G_{AB,B}$ and $G_{A,AB}$

*$G_{A,B}$ is isomorphic to $G_{A^{-1}, B}$ and $G_{A,B^{-1}}$.

*$G_{A,B}$ is isomorphic to $G_{C^{-1}AC, C^{-1}BC}$ for any invertible integer matrix $C$.

I know that all the matrices $B(a,b,c,d)$ are conjugate to $C_4\oplus C_4=\begin{pmatrix} 0&-1\\1&0\end{pmatrix}\oplus \begin{pmatrix}0&-1\\1&0\end{pmatrix}$ but when I conjugate, I move the matrix $A$ and there still are parameters around, and also is it too difficult (I think impossible in general) to find which matrix $P$ makes $B(a,b,c,d)$ similar to $C_4\oplus C_4$ for general $a,b,c,d$.
The presentation of the group $G_{A,B(a,b,c,d)}$ is (computed in Magma calculator):

G1 :=  Group<x1,x2,x3,x4,e1,e2 | (x1,x2), (x1,x3), (x1,x4), (x2,x3), (x2,x4),
(x3,x4), (e1,e2),  x1^(e1)=x1, x2^(e1)=x2^-1, x3^(e1)=x3, x4^(e1)=x3*x4^-1
, x1^(e2)=x1^(-2a-b)*x2^(-(1+(2a+b)^2)/(2c)), x2^(e2)=x1^(2c)*x2^(2a+b),   x3^(e2)=x2^(d)*x3^(-b)*x4^(-2d), x4^(e2)=x1^(c)*x2^(a)*x3^((1+b^2)/(2d))*x4^b>;

Any help will be very appreciated!
 A: Finally I've come with an answer to my question. I wasn't sure about answer because it changes a little bit the approach of the question but I'll do anyway.
The problem originally was about finding a classification (up to isomorphism) about the groups $G_{A,B}$ where $A=P^{-1} \tilde{A} P$ and $B=P^{-1} \tilde{B} P$ are integer matrices and $P\in \mathsf{GL}_4(\mathbb{R})$ and $$\tilde{A}=\begin{pmatrix} 1&0&0&0\\ 0&1&0&0 \\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix}, \quad \tilde{B}=\begin{pmatrix} 0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}.$$
I was dealing with the case $A$ as in the question. Then, as $\tilde{B}$ commutes with $\tilde{A}$, $B$ must commute with $A$ so we can write (also using that $B$ will be of the form $B=\begin{pmatrix} S&*\\0&T\end{pmatrix}$ with $S^4=T^4=I$) $B=\begin{pmatrix} a& -\frac{1+a^2}{b}&0&-\frac{1+a^2}{2b}\\ b&-a & -\frac{d}{2}& \frac{c-a}{2} \\ 0&0&c&-\frac{1+c^2}{d}\\0&0&d&-c\end{pmatrix}$ (I cleared the equations somewhat different as in the question).
Now all that I've to do is find $C\in\mathsf{GL}_4(\mathbb{Z})$ such that $AC=CA$ and $B(a,b,c,d)C=C B(1,1,1,2)$.
First, as $AC=CA$, $C$ must be of the form $C=\begin{pmatrix} p_1&p_3&0&\frac{p_3}{2}\\ p_2&p_4& -\frac{p_8}{2}& \frac{p_4-p_{12}}{2}\\ 0&0&p_7&p_{11}\\ 0&0&p_8&p_{12}\end{pmatrix}$. As $\begin{pmatrix} p_1&p_3\\ p_2&p_4 \end{pmatrix}$ must conjugate $\begin{pmatrix} a& -\frac{1+a^2}{b}\\b&-a\end{pmatrix}$ to $\begin{pmatrix} 1&-2\\1&-1\end{pmatrix}$ we can put $p_3, p_4$ in terms of $p_1$ and $p_2$ (Note that $p_1, p_2, p_3, p_4$ exists because the two matrices are indeed conjugate) and the same for the matrix below. The condition of the determinant being equal to 1 ensures that $p_3, p_8$ and $p_4-p_{12}$ are even and it is straightforward to check that $C$ works so we're done.
