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If $I = \mathbb Z/2\mathbb Z$ counts as a finite $\infty$ category (Edit: it doesn't), then the answer is no. 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 $* \sqcup *$.

If $I = \mathbb Z/2\mathbb Z$ counts as a finite $\infty$ category (Edit: it doesn't), then the answer is no. 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 $* \sqcup *$.

If $I = \mathbb Z/2\mathbb Z$ counts as a finite $\infty$ category, then the answer is no. 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 $* \sqcup *$.

If $I = \mathbb Z/2\mathbb Z$ counts as a finite $\infty$ category (Edit: it doesn't), then the answer is no. 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 $* \sqcup *$.

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 $* \sqcup *$.

If $I = \mathbb Z/2\mathbb Z$ counts as a finite $\infty$ category, then the answer is no. 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 $* \sqcup *$.