= Solution
The <contraction mapping theorem> says that a self-map $g$ of a nonempty <complete metric space> satisfying $d(gx,gy)\leq k d(x,y)$ for a constant $0\leq k<1$ has exactly one <fixed point>. Apply it to $g=f^N$ to obtain $p$ with $f^Np=p$.
No continuity of $f$ itself is needed: composition gives
$$
f^N(fp)=f(f^Np)=fp.
$$
Thus $fp$ 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 $p$. \b[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.
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