Thanks to the posts 65876
and 66141, I was thinking about the functional square roots of the identity function on $\mathbb R$, namely involutions. An involution on $\mathbb R$ is a function $f:\mathbb R \to \mathbb R$ such that $f(f(x))=x$ for all $x \in \mathbb R$. I will restrict myself to functions that are continuous everywhere.
It is an easy exercise that every continuous involution on $\mathbb R$ has at least one fixed point. After a little effort, I have a complete characterization of continuous involutions on $\mathbb R$ with precisely one fixed point. Namely, these are precisely the functions that of the form:
$$
f(x) =
\begin{cases}
a + \varphi(x-a) & x \geq a,
\\
a – \varphi^{-1}(a-x) & x \leq a.
\end{cases}
$$
where $a \in \mathbb R$ and $\varphi: [0,\infty) \to [0, \infty)$ is a continuous strictly increasing bijection with $\varphi(0)=0$. Here $a$ is the fixed point of $f$. (A special example is the function $x \mapsto 2a-x$ for a fixed $a \in \mathbb R$.)
I want to know if there are any nontrivial examples other than the above family of functions.
Are there continuous involutions on $\mathbb R$, other than the identity, that have at least two fixed points?
Note. Searching on Math.SE, I found this question on involutions of $\mathbb R$. Gerry's answer gives a reference to a paper by J F Ritt, On certain real solutions of Babbage's functional equation, Annals of Math 17 (1916) 113-122. The paper touches upon fixed points, but does not address or answer my question.
Best Answer
A continuous involution of $\mathbb{R}$ is either order-preserving or order-reversing, since it's bijective. If it's order-preserving it's not hard to show that it's the identity, and if it's order-reversing it has exactly one fixed point (since it flips the half line above any fixed point with the half line below it).