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Perhaps this is a foolish question, forgive my lack of knowledge of topology. My question is, does there exist a homotopy equivalence from $\mathbb{C}P^{2n}$ to itself that reverses orientation?

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No, there is not. The cohomology ring $H^*(\mathbb{C}P^{N})$ is a truncated polynomial ring $\mathbb{Z}[x]/(x^{N+1})$, where $x$ has degree $2$. Given any map $f:\mathbb{C}P^{N}\to\mathbb{C}P^{N}$, there is some $d\in\mathbb{Z}$ such that $f^*(x)=dx$, and then $f^*(x^N)=d^Nx^N$. If $f$ is a homotopy equivalence, then $d=\pm 1$, and if $N$ is even, this implies $f$ induces the identity on $H^{2N}(\mathbb{C}P^N)$ and thus preserves the orientation. (Note that for $N$ odd, $[z_0,\dots,z_N]\mapsto [\bar{z}_0,\dots,\bar{z}_N]$ is an orientation-reversing diffeomorphism $\mathbb{C}P^N\to\mathbb{C}P^N$.)

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