If $G$ is linear over $\mathbb{Z}$ then $\operatorname{Aut}(G)$ is linear over $\mathbb{Z}$ as well?

Question

If $G$ is linear over $\mathbb{Z}$ then $\operatorname{Aut}(G)$ is linear over $\mathbb{Z}$ as well?

We have $G \cong H \leq GL(n, \mathbb{Z})$ for some $n$ (this is what I mean when I say that $G$ is linear over $\mathbb{Z}$)

Then is $\operatorname{Aut}(G) \cong N \leq GL(n^2, \mathbb{Z})$?

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I need this in context of:

$G$ polycyclic, $\Gamma \leq \operatorname{Aut}(G)$ solvable $\implies \Gamma$ polycyclic

If the topic assertion is correct then this can proven as:

Every polycyclic group is linear of $\mathbb{Z}$, and every solvable linear group of $\mathbb{Z}$ is polycyclic.

Answer 1

The answer to your question is no, because all free groups $F_n$ of finite rank are linear over ${\mathbb Z}$, but it is proved here that ${\rm Aut}(F_n)$ is not linear (over any field) for $n \ge 3$.

It is proved in Chapter 2 of Dan Segal's book on polycyclic groups that solvable subgroups of ${\rm GL}(n,{\mathbb Z})$ are polycyclic.