Assume $0\in Fix(p)$ and $U$ is a small neighborhood of $0$. Let $f(x)=x-p(x)$. Identifying $\mathbb R^n$ with $T_0\mathbb R^n$, set $\pi_{\mathrm{ker}(dp(0))}$ to be the projection $\mathbb R^n\to\mathrm{ker}(dp(0))$. Then $g=\pi_{\mathrm{ker}(dp(0))}\circ f$ is smooth and regular at $0$, so $M=g^{-1}(0)$ is a manifold of dimension equal to the rank of $dp(0)$. By construction, $M$ contains $Fix(p)$. On the other hand, since $Fix(p)$ is the image of $p$, $p$ preserves $M$. Thus, $p|_M$ is a smooth retraction of full rank on $M$. In particular, it's a local diffeomorphism (since it has full rank), and since it's a retraction and we're working in a small neighborood of $0$, it's the identity.

Assume that:

$$0\in Fix(p)$$

and $U$ is a small neighborhood of $0$. 

Let $$f(x)=x-p(x)$$

Identifying $\mathbb R^n$ with $T_0\mathbb R^n$, set $\pi_{\mathrm{ker}(dp(0))}$ to be the projection $\mathbb R^n\to\mathrm{ker}(dp(0))$. 

Then $g=\pi_{\mathrm{ker}(dp(0))}\circ f$ is smooth and regular at $0$, so $M=g^{-1}(0)$ is a manifold of dimension equal to the rank of $dp(0)$. 

By construction, $M$ contains $Fix(p)$. 

On the other hand, since $Fix(p)$ is the image of $p$, $p$ preserves $M$. 

Thus, $p|_M$ is a smooth retraction of full rank on $M$. In particular, it's a local diffeomorphism (since it has full rank), and since it's a retraction and we're working in a small neighborhood of $0$, it's the identity.

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.

Assume $0\in Fix(p)$ and $U$ is a small neighborhood of $0$. Let $f(x)=x-p(x)$. Identifying $\mathbb R^n$ with $T_0\mathbb R^n$, set $\pi_{\mathrm{ker}(dp(0))}$ to be the projection $\mathbb R^n\to\mathrm{ker}(dp(0))$. Then $g=\pi_{\mathrm{ker}(dp(0))}\circ f$ is smooth and regular at $0$, so $M=g^{-1}(0)$ is a manifold of dimension equal to the rank of $dp(0)$. By construction, $M$ contains $Fix(p)$. On the other hand, since $Fix(p)$ is the image of $p$, $p$ preserves $M$. Thus, $p|_M$ is a smooth retraction of full rank on $M$. In particular, it's a local diffeomorphism (since it has full rank), and since it's a retraction and we're working in a small neighborood of $0$, it's the identity.