$f(f(f(x)))=x,$ but $f(x)\neq x$

Let $g:\mathbb{R^1}\to\mathbb{R^2}$ be a one-to-one mapping (which exists, because both sets have the same cardinality). Now set $f=g^{-1}\circ F\circ g:\mathbb{R}^1\to\mathbb{R}^1$, where $F$ is the rotation by $120^\circ$. So the result is not true: this is the desired counterexample.

Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x \not\in \{0,1,2\}$.

You can make $f$ act as a 3-cyclic permutation

$ f(x) = \begin{cases} 0 & \text{if $x = -1$ } \\ 1 &\text{if $x = 0$} \\ -1 &\text{if $x = 1$} \\ x &\text{otherwise} \end{cases}$

I suppose it's also possible to make a continuous version of this.