If $f^N$ is contraction function, show that $f$ has precisely one fixed point. [duplicate]
$f^N$ has one 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 $f(\alpha)=\alpha$.