Let $G = \mathbb{Z}/2\mathbb{Z}$, and consider a short exact sequence of \emph{free} $\mathbb Z$-modules of finite rank endowed with a $G$-action
$$ 0\to \mathbb Z^n \to \mathbb Z^m \to \mathbb Z^k\to 0,$$
and suppose that the maps are also $G$-equivariant. Clearly the sequence splits as sequence of $\mathbb Z$-modules, the question is:
does the SES split equivariantly?
Since, using the representation theory of $G$, any irreducible representation over $\mathbb Z^n$ has rank at most $2$, then I think that we can reduce to the case when $k=2$.
However I expect this to be well known to algebraists.
No. Consider $0 \to \mathbb Z\to \mathbb Z\oplus \mathbb Z\to \mathbb Z\to 0$
Here the action is trivial on the first $\mathbb Z$, permutation of coordinated on the second, and the multiplication on $-1$ on the third. The maps are $f_1(a)=(a,a), f_2(a,b)=a-b$ The reason is that the second $\mathbb Z\oplus \mathbb Z$ is not sum of two $1$-dimensianal representations.