If f is a mapping of a complete metric space $(X,d)$ into itself and $f^N$ is a contraction mapping for some positive integer $N$, then $f$ has precisely one fixed point.
The Banach fixed point theorem is already proven so we can show the following:
$f^N$ has a fixed point $\alpha$, then $f^N(\alpha)=\alpha$ and $f^{N+1}(\alpha)=f(\alpha)$ so $f^N(f(\alpha))=f(\alpha)$, hence $f(\alpha)$ is also a fixed point. By uniqueness we have $f(\alpha)=\alpha$
However, this only shows that $f^N$ has a unique fixed point because the Banach theorem states that the contraction mapping has a fixed point. $f^N$ is a contraction mapping but f is assumed to be not a contraction mapping. How can we show that $f$ also has a unique fixed point?