How many continuous involutions on $\mathbb R$ are there? An involution is a function that satisfies the following: $f = f^{-1}$
MY question is how many involutions can you find in the set of real functions, and how would you go about solving that problem?
EDIT: $f$ does have to be in the family of continuous functions.
 A: There are at least infinitely (in particular uncountably many, with cardinality $\mathfrak c$) many. In particular, for any $a\in \Bbb R$, the function 
$$
f(x)=a-x
$$
is a continuous involution
A: A classification:
First, a continuous bijection $\mathbb R\to\mathbb R$ is either strictly increasing or strictly decreasing.
Case 1. $f$ is strictly increasing.  Let $x\in\mathbb R$ and suppose $x < f(x)$.  Then, by hypothesis, $f(x) < f(f(x)) = x$; contradiction.  Similarly if $x > f(x)$.  Therefore $x = f(x)$.  This being true for all $x$, there is just one such involution: the identity.
Case 2. $f$ is strictly decreasing.
Suppose for contradiction that $f(x) > x$ for all $x$.  But then, in particular, $x = f(f(x)) > f(x)$; contradiction.  So there exists $x_1$ with $f(x_1)\le x_1$.  Likewise there exists $x_2$ with $f(x_2)\ge x_2$.  By continuity, $f$ has a fixed point between $x_1$ and $x_2$, say, $f(x_0) = x_0$.
Subcase 2a. There exist $x>x_0$ with $f(x)>x_0$ and there exist $x>x_0$ with $f(x)<x_0$.  Impossible: by continuity, some $x>x_0$ would have $f(x)=x_0$, so $f$ would not be a bijection.
Subcase 2b.  All $x>x_0$ have $f(x)>x_0$.  But then $f(x_0+1)>x_0=f(x_0)$, contradicting that $f$ is strictly decreasing.
Subcase 2c.  All $x>x_0$ have $f(x)<x_0$.  Define $g\colon[x_0,\infty)\to[x_0,\infty)$ by $g(x) = 2x_0 - f(x)$.  Then $g$ is a continuous bijection.  Conversely (I leave the details to you), any continuous bijection $g\colon[x_0,\infty)\to[x_0,\infty)$ yields a corresponding strictly decreasing continuous involution $f$.
A: Given two real numbers, $b,\theta$  with $\theta\in [0,2\pi)$ I will describe a continuous function $f_{b,\theta}$ which is an involution. 
Instead of giving a formula I'll describe it geometrically.
Draw a $V$ with vertex at the point $(b,0)$ in the $x$-axis. The right and left arms of $V$ make angle $\theta$ and $-\theta$ respectively with the $x$-axis. Now every horizontal line above the $x$-axis will intersect this $V$ at two points. The function simply swaps the $x$ co-ordinates of these two points. Hope this explains it.
