A question arises when we are dealing with a function that is not known to be contraction but the composites of that function to itself is a contraction (it is called eventually contractive). More precisely, consider the following problem:
Let $(X, d)$ be a complete metric space and let $f : X → X$. Suppose that for some integer $p ≥ 1$, the function $g = f^{op}$ ($f$ composite to itself $p$ times) is a contraction, i.e., $f$ is eventually contractive. Can we show that $f$ has a unique fixed point $\overline{x}$.
Yes, we can. Let $\bar{x}$ be the unique fixed point of $f^p$. Then, we note that:
$f^p(f^1(\bar{x}))=f^1(f^p(\bar{x}))=f^1(\bar{x})$.
This implies that $f^1(\bar{x})$ is a fixed point of $f^p$. However, we already know that $\bar{x}$ is the unique fixed point, implying that $f^1(\bar{x})=\bar{x}$. This, finally, implies that $\bar{x}$ is a fixed point of $f^1$, too.
Note: The trick is the uniqueness of the fixed point $\bar{x}$ of $f^p$, following from the Banach fixed point theorem.
Edit: Now, we have to show uniqueness. For this, we assume there exists a $\hat{x}\neq\bar{x}$, such that $f^1(\hat{x})=\hat{x}$. We can directly conclude that also $f^p(\hat{x})=\hat{x}$, i.e. $\hat{x}\neq\bar{x}$ must also be a fixed point of $f^p$. This is a contradiction, since $\bar{x}$ is the unique fixed point of $f^p$.