Suppose $M \in GL_k(\mathbb{Z})$ is of order $2$. That is, $M^2 = 1$ and $M \ne 1$. Then is it true that upto a change of $\mathbb{Z}$ basis, $M$ has the form $$\begin{pmatrix}J \\ & J \\ & & \ddots\\ & & & J\\ & & & & \pm 1\\ & & & & & \ddots\\ & & & & & & \pm 1\end{pmatrix}$$ where $J$ is the $2 \times 2$ matrix $\begin{pmatrix}0 & 1\\ 1 & 0 \end{pmatrix}$?
Note that since the minimal polynomial of $M$ divides $x^2-1$, $M$ is diagonalizable over $\mathbb{C}$, and has only $\pm 1$ as eigenvalues. So $M$ is similar to such a matrix over $\mathbb{C}$. From Similar matrices and field extensions it follows that they are similar over $\mathbb{Q}$. It is also not very difficult to show the similarity over $\mathbb{Q}$ directly. Now the statement in some sense reduces to 'the only obstruction to diagonalization of $M$ over $\mathbb{Z}$ is that $J$ cannot be diagonalized'.
From a representation point of view, this is asking for a decomposition of a class of finitely generated modules over $R = \mathbb{Z}[\mathbb{Z}_2]$. Let $\mathbb{Z}_2 = \{1,t\}$, $\mathbb{Z}_+$ stand for the trivial module structure on $\mathbb{Z}$, and $\mathbb{Z}_-$ stand for the module given by $t\cdot n = -n$. Then the above statement is equivalent to saying that every finitely generated module over $R$ which is free over $\mathbb{Z}$ can be decomposed as a direct sum of the irreducibles $R, \mathbb{Z}_+, \mathbb{Z}_-$.
