Continuous involutions on $\mathbb R$ with at least two fixed points 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. 
 A: 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).
A: None other than the identity.
If $f(f(x))=x$, then $f$ is surjective.  Also, $f$ is injective since $f(a)=f(b)\Rightarrow f(f(a))=f(f(b))\Rightarrow a=b$.  So $f$ is invertible.
Now you have that $f(x)=f^{-1}(x)$.  The graph of $f$ must be symmetric about the line $y=x$.  
Suppose that $a$ and $b$ are two fixed points discretely separated with $a<b$ and no fixed points in between.  The graph of $f$ either lies entirely below $y=x$ or above it on the interval $(a,b)$.  (Otherwise, the intermediate value theorem gives another fixed point inbetween.) This contradicts the symmetry of the graph.  (If say this section of the graph lies below the diagonal, then on the same interval, the graph of $f^{-1}$ lies above the diagonal.)
So any example with more than one fixed point will have an entire interval $[a,b]$ where $f(x)=x$.  If the graph ever deviates from the identity function (say after $x=b$) then the graph either moves above or below $y=x$, again contradicting the symmetry.
