Let $G$ be a finite group and $X, Y$ be "nice" free topological G-spaces (such as finite free G-simplicial complexes). Moreover assume $f, g: X\to Y$ are $G$-maps. I have a feeling that the following assertion might be true:
$f$ and $g$ are G-homotopy equivalent iff they are ordinary homotopy equivalent.
Is this observation true? If yes, then is it a well-known fact or is it easy to construct the G-homotopy between them by knowing that they are just homotopic?
Comment I: This intuition comes from the following fact: Given a finite group $G$ and a G-topological space X, then any two G-maps from X to $EG=G\ast G\ast\cdots $ (the total space of the universal bundle of G) are $G$-homotopic. Note that $EG$ is a contractible space so any two maps (from the same domain) to these spaces are ordinary homotopic.
Comment II: Also, there is another piece of evidence for this intuition. There is a famous theorem by Bredon which says
A map $f: X\to Y$ between $G$-CW-complexes is a $G$-equivariant homotopy equivalence iff $f^{H}: X^H\to Y^H$ is a homotopy equivalence for all subgroups H, where in above $X^H= \{x : hx=x\,\, \forall h\in H\}$ and $f^{H}$ is the restriction of $f$ to $X^H$.
Note that if the action is free then in particular this theorem says:
$f$ is G-homotopy equivalent iff it is ordinary homotopy equivalent.
(I should admit that I do not know the proof of this theorem as well.)
Edited later: Let me ask a weaker version of my question here. Let $X$ be a finite free $\mathbb{Z}_2$-simplicial complex, and $f,g : X\to \mathbb{S}^n$ are $\mathbb{Z}_2$-equivariant maps. Moreover, assume $n > \dim(X)$. Is it true that $f$ and $g$ are $\mathbb{Z}_2$-homotpic?
(Note that any two such a map are ordinary homotopic as such maps are always null-homotopic.)