Let $G$ be a finite group, and $X$ be a free $G$-space. Moreover, assume that $X$ has a homotopy type of a CW-complex. Does $X$ have $G$-homotopy type of a $G$-CW complex also?


**Edit:** My main motivation for this question is as follows. Milnor has shown that if $Y$ has a homotopy type of a CW-complex and X is a compact Hausdorff space, then the space of all continuous map $M(X, Y)$ from $X$ to $Y$, endowed with the compact-open topology has a homotopy type of a CW-complex as well. Now, let G be a finite group, and $X$ and $Y$ be a G-spaces. We can equip M(X, Y) to the $G$-action $(g, f)\to g\cdot f$ where $(g\cdot f)(x)=g\cdot f(g^{-1}x)$ for all $x\in X$. I am interested to know whether M(X, Y) has the homotopy type of a $G$-CW complex for nice $G$- spaces, i.e, $X$, and $Y$ are finite G-CW-complexes. Moreover, I am interested in particular in the case that this action is free. For example, this action is free on $M(\mathbb{S}^n, \mathbb{S}^k)$ where $n > k$, as by Borsuk-Ulam theorem there is no $\mathbb{Z}\_2$-map form a higher dimensional sphere to a lower dimensional sphere (Here, we consider $\mathbb{S}^m$ as a free $\mathbb{Z}\_2$-space with the antipodal action) 



Homotopy type of G-CW-structure
There is a paper by Stefan Waner from 1980, I think it's called "Equivariant Classifying Spaces", in which he proves an equivariant version of Milnor's theorem. It might do what you want.


