Prove or disprove that $f$ is identity function given that $f$ is continuous and $f(f(f(x)))=x$

Question

Let $f $ be continuous on $\mathbb{R}$ satisfying $f(f(f(x)))=x$ for all $x\in \mathbb{R}$. Prove/disprove that $f$ is an identity function. So far I have figured out that any point $x$ such that $f(x)\neq x $ is not isolated(if it exists). i.e every neighbourhood of such a point contains other points where function is not identity. Cant proceed any further!

Answer 1

HINT: Using Sharkovsky's theorem:

Proof: Assume $f(x_0)\neq x_0$ for some $x_0\in\Bbb R$. This, along with the condition that $(f\circ f\circ f)(x_0)=x_0$, means that $x_0$ is a periodic point of least period $3$. We then use Sharkovsky's theorem, which states that: "If a continuous real function has a periodic point with least period $3$, then there is a periodic point of least period $n$ for every integer $n$." In particular, there should be a periodic point of least period $5$. But every point is $3$-periodic, so we have a contradiction and conclude that $f(x)=x$ everywhere.