**EDIT**: I've made a mistake with the matrices. Now it is corrected.

A couple of days ago I asked this question. There, answerers gave me excellent hints to solve that case and others too. But I've found two matrices for which I had to distinguish the corresponding groups and I couldn't solve the problem with any of those techniques (see below).

I'm almost done with my task of analyzing these matrices and groups and I think the following are the last examples which I've to distinguish.

Let $A=\begin{pmatrix} 1&0&0&0&0\\0&0&-1&0&0 \\ 0&1&-1&0&0\\ 0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus A'$ and $B=\begin{pmatrix} 1&0&0&0&0\\ 0&0&-1&1&0\\0&1&-1&0&0\\0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus B'$.

Question: Are isomorphic $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^5$ and $G_B=\mathbb{Z}\ltimes_B\mathbb{Z}^5$? Well again I think they're not.

**Thoughts and advances**:

$\bullet$ $B$ is not conjugate to $A$ or $A^{-1}$ in $\mathsf{GL}_5(\mathbb{Z})$ but they are in $\mathsf{GL}_5(\mathbb{Q})$. They are both of order 6 and have 1 as eigenvalue.

$\bullet$ I computed the 2 and 3 exponencial central classes up to 11 (as the answerers taught me in the previous question) and result in isomorphic pQuotients. The presentations are:

```
> GA := Group<a,b,c,d,e,t | (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e),
> a^t=a, b^t=b^-1*c^-1, c^t=b, d^t=d*e^-1, e^t=d>;
>
> GB := Group<a,b,c,d,e,t | (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e),
> a^t=a, b^t=b^-1*c^-1, c^t=b, d^t=b*c*d*e^-1, e^t=b*c*d>;
```

$\bullet$ I've found in this paper Corollary 8.9 (cf Prop 4.2 and Def 4.3) that if I had $\mathbb{Z}\ltimes_{A'} \mathbb{Z}^4$ and $\mathbb{Z}\ltimes_{B'}\mathbb{Z}^4$ then those semidirect products wouldn't be isomorphic because $B'\not\sim A',(A')^{-1}$ in $\mathsf{GL}_5(\mathbb{Z})$ (and because neither have 1 as eigenvalue) but I don't know how to relate these semidirect products with the original ones I have.

$\bullet$ $G_A^{ab}\cong G_B^{ab}\cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_3$. Also I tried to compute the quotients $G/\gamma_i(G)$ (for $i\geq 2$) where $\gamma_i=[\gamma_{i-1}(G),G]$ and $\gamma_1=[G,G]$ and all of them are isomorphic.

$\bullet$ Thinking of $\Gamma_A=(G_A/Z(G_A))$ and $\Gamma_B=(G_B/Z(G_B))$ I get $\Gamma_A\cong \mathbb{Z}_6\ltimes_{A'}\mathbb{Z}^4$ and $\Gamma_B\cong \mathbb{Z}_6\ltimes_{B'}\mathbb{Z}^4$ and I computed the abelianization ($\mathbb{Z}_6\oplus\mathbb{Z}_3$) and pQuotients here too but I couldn't distinguish them either.

```
> Gamma_A := Group<a,b,c,d,t | (a,b), (a,c), (a,d), (b,c), (b,d),
> (c,d), t^6, a^t=a^-1*b^-1, b^t=a, c^t=c*d^-1, d^t=c>;
>
> Gamma_B := Group<a,b,c,d,t | (a,b), (a,c), (a,d), (b,c), (b,d),
> (c,d), t^6, a^t=a^-1*b^-1, b^t=a, c^t=a*b*c*d^-1, d^t=a*b*c>;
```

I hope someone can help me again with this.