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$S^3 \times \mathbb{R}P^2$ and $\mathbb{R}P^3 \times S^2$ are both smooth 5-manifolds with fundamental group $\mathbb{Z}/2$ and universal cover $S^3 \times S^2$, so their homotopy groups are all the same. On the other hand, only the latter is orientable since $\mathbb{R}P^3$ is orientable but $\mathbb{R}P^2$ isn't, so they have different values on $H^5$ and therefore can't be homotopy equivalent. (I think this example is in Hatcher somewhere.)