 Solution

The contraction mapping theorem says that a self-map $g$ of a nonempty complete metric space satisfying $d(gx,gy)\le kd(x,y)$ for a constant $0\le k<1$ has exactly one fixed point. Apply it to $g=f^N$ to obtain $p$ with $f^{N}p=p$.

No continuity of $f$ itself is needed: composition givesThus $fp$ is also a fixed point of $f^N$, whose uniqueness forces $fp=p$. Conversely any fixed point of $f$ is fixed by $f^N$, so it must equal $p$. The unique fixed point of the contracting iterate is also the unique fixed point of $f$. The usual nonempty-space hypothesis is necessary; an empty space has no fixed point.