No, let  $I = \mathbb Z/2\mathbb Z.$  The space $* \sqcup *$  admits both a free and a trivial $I$ action.  We have an additive map given by taking homology and pairing with the alternating representation $$X \mapsto \chi(H^*(X, \mathbb C), {\rm alt}) = \sum_{i} (-1)^i \dim (H^i(X, \mathbb C) \otimes{ \rm alt})^{\mathbb Z/2}$$  which distinguishes between the two actions on $* \sqcup *$.