Iterating a real continuous injective function having no fixed points.

Let $f: \mathbb R \rightarrow\mathbb R$ be a continuous injective function. If $f(x)≠x ,\forall x∈\mathbb R$ and there exists a positive integer $n$ such that $f^n(x)=x , \forall x∈\mathbb R$ , then how do we prove that $f^2(x)=x ,\forall x∈\mathbb R$ ?

Answer

I don't think any such $f$ exists, hence the statement is vacuously true.

Since $f$ is injective, it must be strictly monotonic.

Since $f(x) \neq x$ for all $x$, and $f$ is continuous, we must have (i) $f(x)x$for all$x$. In the first case, we see that$\lim_{x \to -\infty} f(x) = -\infty$, and in the second case,$\lim_{x \to \infty} f(x) = \infty$, hence$f$ must be increasing.

Combining, we see that $f$ is strictly increasing.

In Case (i), $f(x) < x$, hence $f(f(x)) < f(x)$, which gives $f^2(x) < x$.

In Case (ii), $f(x) > x$, hence $f(f(x)) > f(x)$, which gives $f^2(x) > x$.