I wonder if $f: X\rightarrow X$ from the metric space $X$ to itself is such that there exists $n>1$ such that $f^n$ is a contraction, then it is true that for any point $x\in X$, $(f^m(x))_m$ is a Cauchy sequence? I think of this question when I was doing an exercise that asks to show that if $X$ is complete, then $f$ has a unique fixed point. I was able to solve this exercise but still curious about the answer of my question.
Thanks in advance!
Since $f^n$ is a contraction, we can find $\lambda\in(0,1)$ such that $d(f^nx,f^ny)\leq\lambda d(x,y)$ for all $x,y\in X$.
Fix $x\in X$, and we can partition $(f^mx)_m$ into $n$ subsequences $(x_k(r)=f^{nk+r}x)_k$ for $r=0,1,\dots,n-1$. We know each $(x_k(r))_k$ is Cauchy because $f^n$ is a contraction. Let $\operatorname{diam}\{f^mx\mid m\geq 0\}=\Delta$ (this is finite because we have finitely many Cauchy sequences). Then \begin{align} \operatorname{diam}\{f^mx\mid m\geq m_0=nk+r\} &\leq\operatorname{diam}\{f^mx\mid m\geq nk\}\\ &\leq\lambda^k\Delta\to 0 \end{align} as $k\to\infty$ (equivalently $m_0\to\infty$), so $(f^mx)_m$ is Cauchy.