Suppose $(X, \rho)$ is a complete metric space, and suppose the function $T : (X, ρ) \rightarrow (X, ρ)$ is such that $T_n = T ◦ T ◦ · · · ◦ T$ (n times) is a contraction map for some $n \ge 2$. Prove that there exists a unique $x^* \in X$ such that $T(x^* ) = x^*$ .
My attempt:
Since $T_n$ is a contraction, the Contraction Mapping Theorem implies that there is a unique $x^* \in X$ such that $T_n(x^*) = x^*$. Then observe that
$$T(x^*) = T(T_n(x^*)) = T_{n+1}(x^*) = T_n(T(x^*)).$$
So now that we have derived that $T(x^*) = T_n(T(x^*))$ can we say $T(x^*) = x^*$ must be true since it's that only solution s.t. $T_n(x^*) = x^*$ or is there more to be said about $T(x^*) = T_n(T(x^*))$ that leads to the hypothesis that: exists a unique $x^* \in X$ such that $T(x^* ) = x^*$.
This is to almost finish your attempt: We had earlier that $T_n(x^*) = x^*$. Now we have $T_n(T(x^*)) = T(x^*)$ i.e. $y = T(x^*)$ is also a solution the the equation $T_n(x) = x$. By uniqueness, $y = x^*$ i.e. $T(x^*) = x^*$.
So yes, we can say $T(x^*) = x^*$ is true by uniqueness guaranteed by the contraction mapping theorem applied to $T_n$.
Note that we haven't shown that $T(x) = x$ has a unique solution! This is the final part of the proof, but this should be easy.
Addendum: final part of the proof.
We had $T(x^*) = x^*$. Now, suppose $\alpha \in X$ is also a solution to $T(x) = x$ i.e. $T(\alpha) = \alpha$. Then
$$T_n(\alpha) = T_{n-1}(T(\alpha)) = T_{n-1}(\alpha) = T_{n-2}(T(\alpha)) = T_{n-2}(\alpha) = \ldots = T(\alpha) = \alpha.$$
Now we have $T_n(\alpha) = \alpha$ and also $T_n(x^*) = x^*$ from earlier. By uniqueness guaranteed by the contraction mapping theorem on $T_n$, we must have $\alpha = x^*$. So indeed, $T(x) = x$ has a unique solution.