Examples of involutions on $\mathbb{R}$

If you assume continuity, you still have as many involutions as continuous functions from $\mathbb R$ to itself, namely cardinality $c$. For if $f$ is any strictly decreasing continuous function from $[0, \infty)$ onto $(-\infty, 0]$, you can extend $f$ to $(-\infty, 0)$ by $f(t) = x$ where $f(x) = t$, and you get a continuous involution.

If we do not put any conditions on $f$, there will be far too many involutions. Divide the reals in any way into a disjoint union of $1$-element sets and/or $2$-element sets.

For any such subdivision $\mathbb{U}$, if $\{a\}$ is a singleton set in the subdivision, let $f(a)=a$. If $\{a,b\}$ is a doubleton set in the subdivision, let $f(a)=b$ and $f(b)=a$. Then $f$ is an involution. Any subdivision determines an involution, and conversely every involution determines a subdivision.

It follows that there are $2^\mathfrak{c}$ involutions, "just as many," from the point of view of cardinality, as the number of functions from $\mathbb{R}$ to $\mathbb{R}$.

The set of continuous involutions on $\mathbb R$ has a fairly easy to describe structure. Let $f$ be a continuous involution on $\mathbb R$. Then:

1. Any point is either fixed, or of period $2$. The set of fixed points is closed, so the set of periodic points is open.
2. If $x$ is of period $2$, there is a fixed point between $x$ and $f(x)$.
3. If $x$ is of period $2$, there is exactly one fixed point between $x$ and $f(x)$. Let $a$ be the nearest fixed point to $x$ between $x$ and $f(x)$. Then as $x$ approaches $a$, $f(x)$ must approach $a$, and can never be equal to a fixed point, since by our choice of $a$, $x$ is never itself fixed as it approaches $a$. This is impossible if there's another fixed point between $x$ and $f(x)$.
4. If $f$ has at least two fixed points, then suppose there is a non-fixed point in between them. Take the largest open interval of non-fixed points around that point. Its boundaries are fixed. By continuity and (2), near the left edge of that interval, $f(x)$ must be outside of the interval on the left, and near the right edge, $f(x)$ must be outside on the right. As $x$ runs through the interval, $f(x)$ can't touch any fixed points, so this violates the intermediate value theorem.
5. If on the other hand, any two fixed points of $f$ have no non-fixed points between them, then the set of fixed points forms a single closed interval. If this interval isn't all of $\mathbb R$, then (3) is impossible to satisfy for any non-fixed $x$.

Thus $f$ either fixes $\mathbb R$ or has exactly one fixed point, dividing the real line into two halves. By (2), the left half must map onto the right half and vice-versa. For simplicity let's say the fixed point is $0$. Then $f$ can/must be equal to any/some strictly decreasing continuous function with $f(0)=0$ on $\mathbb R^+$, and to its inverse on $\mathbb R^-$. For instance we can take $f(x)=-2x$ for positive $x$ and $f(x)=-\frac12x$ for negative $x$.