This should probably be a comment, but I lost my old account and can't yet make comments.

Assume $0\in Fix(p)$ and $U$ is a small neighborhood of $0$. Let $f(x)=x-p(x)$. Then $g=\pi_{\mathrm{ker}(dp(0))}\circ f$ is smooth and regular at $0$, so $Fix(p)$ is contained in a manifold $M=g^{-1}(0)$ of dimension equal to the rank of $dp(0)$. On the other hand, $p$ fixes $M$, so it is a smooth retraction of full rank on $M$, so it must be the identity.