A facetious answer would be: sure, but one of the ingredients is going to be the group of homotopy equivalences between $X^G$ and $Y^G$. Since $G$-spaces are diagrams of fixed spaces, the Dwyer-Kan obstruction theory of diagrams (c 1984) probably describes the space of maps, in principle allowing you to answer this question and decide which maps of the given fixed sets are realized. This doesn't sound remotely practical to me.

If you don't understand the fixed points to begin with, Smith theory or bust. The map on fixed sets is going to be an isomorphism on homology with coefficients (possibley $0$) depending on the group and exotic fixed sets satisfying that condition are possible. If you're willing to assume everything is simply connected and the group is the circle, it's probably automatically an equivalence. On the opposite extreme is (binary?) $A_5$ acting on high dimensional manifolds. It has PL actions on the disk with the fixed set arbitrary simplicial complexes. I forget what happens in the smooth category; I think empty fixed set is possible, but I forget how. In between are $p$-groups, where you'll keep control of the mod $p$ homology but not the general homology.