Invariant subgroups of $\mathbb{Z}^m$ and commutativity

Question

$\DeclareMathOperator\Im{Im}$Given a unimodular matrix $A$ (of finite order).
Every finite index subgroup of $\mathbb{Z}^m$ can be written as $\Im B = B\mathbb{Z}^m$ for some square integral matrix $B$. (In fact, by the Hermite normal form over the integers, we may assume $B$ is upper triangular with non-zero diagonal.)

I would like to show that there exists a fixed number $M$, such that if $\Im AB = \Im B$, i.e. the subgroup $\Im B$ is invariant under $A$, then there exists some matrix $C \in \mathbb{Z}^{m\times m}$ with $0\neq |\det C| \leq M$ for which $BC$ commutes with $A$.
In some sense, I wish to show that the condition $\Im AB = \Im B$ is 'close' to commuting. Indeed, we can then replace $\Im B$ by a (not too much, see the bound $M$) smaller subgroup contained in it that allows a matrix notation that commutes with $A$. It is necessary for $M$ to be independent of $B$.

My current idea goes as follows:
I would like to show that $G = \{B \in \text{GL}(m,\mathbb{Q}) \mid \Im AB = \Im B\}$ is a subgroup of $ \text{GL}(m,\mathbb{Q})$. Clearly, $C_{\text{GL}(m,\mathbb{Q})}(A) = \{B \in \text{GL}(m,\mathbb{Q}) \mid AB = BA\}$ is also a subgroup and $C_{\text{GL}(m,\mathbb{Q})}(A)$ is a subgroup of $G$ (if the first assertion holds). Secondly, I would like to show that $C_{\text{GL}(m,\mathbb{Q})}(A)$ has finite index in $G$.
Suppose these two things hold. Let $C_1$ to $C_k$ be representatives of the cosets. By taking scalar multiples over $\mathbb{Q}$, we can assume all $C_i^{-1}$ are integral. Take $M = \max\{|\det C_i^{-1}|\}$. If $B$ is an integral matrix of $G$ that lies in the coset of $C_j$, then $BC_j^{-1}$ commutes with $A$. This finishes the proof.

I am however stuck trying to show the two assertions about $G$ being a group and $C_{\text{GL}(m,\mathbb{Q})}(A)$ having finite index in $G$. Other approaches and/or a link to good literature are also welcome.
Can we also do this when we replace $A$ by an integral representation of a finite group (hence the choice of a unimodular matrix of finite order in my question)?

Answer 1

The answer to the first part goes as follows:
Let us do it in general for an integral representation of a finite group. We have given $B$ satisfying $\text{Im}\rho(g)B = \text{Im}B$ for all $g\in G$ and $\rho: G \to \text{GL}(m,\mathbb{Z})$. This condition implies that $\rho(g)B = B\sigma(g)$ for some integral representation $\sigma: G \to \text{GL}(m,\mathbb{Z})$. In other words, $\rho$, $\sigma$ are $\mathbb{Q}$-equivalent integral representations. The set $\{\sigma\mid \sigma \sim_\mathbb{Q} \rho\}$ splits in a finite number of $\mathbb{Z}$-equivalence classes.
Let $\sigma_1$ to $\sigma_k$ be representatives of theses classes. By definition, we get matrices $C_i\in\text{GL}(m,\mathbb{Q})$ such that $\rho(g)C_i = C_i\sigma_i(g)$. We can choose $C_i$ in such a way that $C_i^{-1} \in \mathbb{Z}^{m\times m}$. Pick $M = \max\{|\det C_i^{-1}|\}$.
Now $\sigma$ must by $\mathbb{Z}$-equivalent to some $\sigma_i$, say $\sigma(g) = P^{-1}\sigma_i(g)P$.
By construction, we have $\rho(g)C_i = C_i\sigma_i(g)$ and $\rho(g)B = B\sigma(g) = BP^{-1}\sigma_i(g)P$. Hence it follows that $\rho(g)BP^{-1}C_i^{-1} = BP^{-1}C_i^{-1}\rho(g)$, and by construction $P^{-1}C_i^{-1}\in \mathbb{Z}^{m\times m}$ with its determinant smaller than $M$.