You don't need to show that $T$ is a contraction. That might be false. (E.g., $X=\mathbb R^2$, $T(x,y)=(0,2x)$, $m=2$.)
then apply the fixed point theorem
So you know a fixed point theorem that would apply if $T$ were a contraction. That means that you know a fixed point theorem that does apply to $T^m$. Hence, you know that $T^m$ has a unique fixed point $x_0\in X$.
- $T$ cannot have any other fixed points, because every fixed point of $T$ is a fixed point for all powers of $T$.
- Thus the remaining work is to show that $x_0$ is in fact also a fixed point for $T$.
Note that $x_0=T^m(x_0)$ and $T(x_0)=T(T^m(x_0))=T^m(T(x_0))$, so
- $d(T(x_0),x_0)=d(T^m(T(x_0)),T^m(x_0))\leq \lambda_1 d(T(x_0),x_0)\implies d(T(x_0),x_0)=0.$
- Alternatively, as noted, $T^m(T(x_0))=T(x_0)$, which shows that $T(x_0)$ is a fixed point for $T^m$, hence $T(x_0)=x_0$ by uniqueness.