Is there a function $f:\mathbb{R}\to\mathbb{R}$ such that $f\circ f\circ f=Id.$, but $f\neq Id.$?

Question

Is there any function $f:\mathbb{R}\to\mathbb{R}$, other than the identity, such that $$f\circ f\circ f=Id.?$$

That is a pretty simple question, but surprisingly I am not able to say anything about it. If we require only $$f\circ f=Id.,$$ then, there are some answers, like $f(x)=1/x$ and $f(x)=-x$, for example. And if we allow $f$ to be over $\mathbb{C}$, then the original question also has some answers, like $f(x)=e^{2\pi i/3}x$. But none of these generalize to $f:\mathbb{R}\to\mathbb{R}$ such that $f^3=Id.$. Do you have any idea?

Answer 1

Without requiring continuity of $f$, the answer is yes. For example, define $f(x) = x+1$ for $3n \le x < 3n+2$ and $x - 2$ for $3n+2 \le x < 3n+3$, for integers $n$.

Answer 2

$$f(x)=\begin{cases}x+2 & \lfloor x \rfloor \equiv 0 \pmod{3}\\ x-1 & \text{ otherwise}\end{cases}$$