Projective G-group Let $G$ be a fixed group. By definition, a $G$-group is a group $X$ with a $G$-action that respects the group operation of $X$. A free $G$-group means a group freely generated by a free $G$-set. A "projective $G$-group" means a projective object in the category of $G$-groups.
Can there be projective $G$-groups which are not free $G$-groups?
If yes, for which groups $G$ does it happen?
 A: No (there do not exist projective $G$-groups which are not free $G$-groups).
By standard arguments, a $G$-group is projective if and only if it is a retract of a free $G$-group. So we only have to show that a retract of a free $G$-group is a free $G$-group.
This is not immediate, because a $G$-subgroup of a free $G$-group need not be a free $G$-group, as noted in the comments. However it follows from considerations of equivariant group cohomology.
Let $H$ be a $G$-group, and let $M$ be a $\mathbb{Z}[H\rtimes G]$-module. Cegarra, García-Calcines and Ortega define the equivariant group cohomology of $H$ with coefficients in $M$ to be $H^*(H\rtimes G,G;M)$, the Takasu relative cohomology of the pair (which agrees with the singular cohomology of the pair $(B(H\rtimes G),BG)$ for suitably chosen classifying spaces). This has the usual functoriality properties for homomorphisms of $G$-groups. Define the equivariant cohomological dimension of the $G$-group $H$ to be the maximal $n$ such that $H^n(H\rtimes G,G;M)\neq 0$ for some coefficient module $M$.
In a recent preprint with Kevin Li, Ehud Meir and Irakli Patchkoria, we prove the following:
Theorem: A $G$-group has equivariant cohomological dimension $\le1$ if and only if it's a free $G$-group.
Now if the identity on $H$ factors as $H\to F\to H$ where $F$ is a free $G$-group, then the equivariant cohomological dimension of $H$ is less than or equal to $1$. Hence projective $G$-groups are free $G$-groups.
(Actually this is just one of three possible definitions of equivariant cohomological dimension, which are compared in the preprint.)
