This question already has an answer here:
If $f$ is a mapping of a complete metric space $(X, d)$ into itself and $f^N$(composite $f$ for $N$ times) is a contraction mapping for some positive integer $N$, then $f$ has precisely one fixed point. (Banach fixed point theorem is applicable)
I tried to show that $f$ is also a contraction function. I considered the sequence $x,f(x),f^2 (x),... $, but then fail pathetically in showing that it's cauchy. Please tell me if my direction is correct, any new ideas are appreciated. Thanks.