If we let $X$ be a complete metric space, and let $S:X\to X$ be a map, such that $S^m$ is a contraction. We now want to show, that $S$ has a unique fixed point
This is what I've thought so far:
Due to Banachs fixed point theorem is it enough to show, that $S$ is a contraction. Due to $S^m$ being a contraction, we know this about $S^m$(the definition of being a contraction): $$\exists\beta, 0\le\beta\lt1:d(S^mx,S^my)\le\beta d(x,y),\forall x,y\in X$$
I'm not really sure how to show that S is a contraction.. Any ideas ad to how to approach this?