Extending scalars from $\mathbb{Z}[G]$ to $\mathbb{Q}[G]$ 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!
 A: 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
