Let $G$ be a finite group.
Let $M$ and $N$ be finitely generated $\mathbb{Z}[G]$-modules such that $M$ is free as a $\mathbb{Z}$-module.
Suppose that $\mathbb{Q}\otimes_\mathbb{Z}M$ and $\mathbb{Q}\otimes_\mathbb{Z}N$ are isomorphic as $\mathbb{Q}[G]$-modules.
I want to show that there exists a $\mathbb{Z}[G]$-homomorphism $\phi:N \to M$ such that the map $$\mathrm{id}_\mathbb{Q}\otimes_\mathbb{Z}\phi: \;\mathbb{Q}\otimes_\mathbb{Z}N \to \mathbb{Q}\otimes_\mathbb{Z}M$$ is an isomorphism of $\mathbb{Q}[G]$-modules.
Any hints very gratefully received! I'm sure I must be missing something simple!
From some of the answers to my comment, it seems that it wasn't very clear, so let me make it clearer. The idea is the one I said : by decomposing $N= Tor(N)\oplus \mathbb{Z}^r$, although this decomposition need not be $G$-equivariant, the projection $N\to \mathbb{Z}^r$ can be made $G$-equivariant, and then by picking the right iso $\mathbb{Q}\otimes N\to \mathbb{Q}\otimes M$ we can find a map $\mathbb{Z}^r\to M$ that induces the iso.
Here are the details : any element of $G$ sends $Tor(N)$ to $Tor(N)$ because they are isomorphisms, therefore it induces a map $N/Tor(N)\to N/Tor(N)$ (its inverse is also an element of $G$ and so induces a map too, so clearly we actually get a $G$-module structure on $N/Tor(N)$). This $G$-structure on $N$ is such that the projection map $N\to N/Tor(N)$ (i) induces an isomorphism when tensored with $\mathbb{Q}$, (ii) is $G$-linear.
Now the structure theorem for finitely generated abelian groups tells us that $N/Tor(N) = \mathbb{Z}^r$, and $M$ is free, and $N\otimes \mathbb{Q}$ is isomorphic to $M\otimes \mathbb{Q}$, therefore $M=\mathbb{Z}^r$.
Now pick an isomorphism $f:\mathbb{Q}\otimes N = \mathbb{Q}^r \to \mathbb{Q}\otimes M = \mathbb{Q}^r$. Multiply it by a sufficiently large integer $n$ so that $f(e_i) \in \mathbb{Z}^r$ for all $i$, where $(e_i)$ is the standard basis. Then $f$ restricts to a map $\mathbb{Z}^r \to \mathbb{Z}^r$, which induces an iso when tensored with $\mathbb{Q}$, and which is $G$-equivariant if you see it as a map $N/Tor(N)\to M$.
Composing this with $N\to N/Tor(N)$ yields the desired map.
Note how we see here that we can't always get a map $M\to N$ : there is no $G$-map $N/Tor(N) \to N$ in general