No: Consider $\mathbb{Z_{2}}$ acting on the universal cover $X=S^{n}$ of $Y=\mathbb{RP}^{n}$ for $n$ odd. In this case both $X$ and $Y$ are closed and orientable, so we can talk about Poincare Duality and transfer maps. The map $\pi^{*}$ must be zero when restricted to the certain degrees of the cohomology groups, as $H^{k}(Y;\mathbb{Z})=\mathbb{Z}_{2}$ and $H^{k}(X;\mathbb{Z})=0$ for $k$ even with $0 < k < n$.
Russell
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