$f(f(f(x)))=x,$ but $f(x)\neq x$ Let $f: \mathbb{R}\to \mathbb{R}$ be strictly increasing function such that  $f(f(f(x)))=x$ then by this $f(x)=x.$ Does the result fail if we drop the hypothesis that f is strictly increasing? 
I find such an example if I change the domain and codomain to $\mathbb{R}^2$ and $f=rotation \ by\  120^o,$ but could not think of a map from $\mathbb{R}\to \mathbb{R}.$ 
Any help is appreciated. 
 A: Define $f(0)=1, f(1)=2, f(2)=0$ and $f(x)=x$ for $x \not\in \{0,1,2\}$.
A: 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.
A: If you allow yourself to consider the projective line (i.e., the reals with a point at infinity), then the function 
$$
f(x) = \frac{1}{1-x}
$$
is a nice continuous solution to the problem (although you have to make sense of "continuous" for this extended line). Pierre Samuel's book on Projective Geometry has a nice exposition of this around page 58. 
(By the way, this answer was inspired by @DonaldSplutterwit's deleted answer, although I might have stumbled on it myself, since I've been thinking about projective geometry and looking at Samuel's book today anyhow. Thanks, Donald!)
A: 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.
