There are a truly enormous number of solutions, if one only wants the solution to work on an interval. Indeed, one can find solutions to f(f(x)) = g(x)$f(f(x)) = g(x)$ for any function g$g$ defined on an interval.
Specifically, I claim that if g:[a,b] to R$g:[a,b] \to \Bbb R$, then there are 2|R|$2^{|\Bbb R|}$ many functions f$f$ from R$\Bbb R$ to R$\Bbb R$ with f(f(x)) = g(x)$f(f(x)) = g(x)$ for all x$x$ in [a,b]$[a,b]$.
One such solution f$f$ is obtained as follows. First choose a z$z$ such that [a,b]$[a,b]$ and [a + z, b + z]$[a + z, b + z]$ are disjoint. Now let f(x) = x + z$f(x) = x + z$, for x$x$ in [a,b]$[a,b]$, and f(x) = g(x - z)$f(x) = g(x - z)$, for x$x$ in [a + z, b + z]$[a + z, b + z]$. Thus, f(x)$f(x)$ first translates x$x$ to another interval, when x$x$ is in [a,b]$[a,b]$, and then f$f$ computes g$g$ of the reverse translate, when x$x$ is not in [a,b]$[a,b]$. So f(f(x)) = g(x)$f(f(x)) = g(x)$.
When g$g$ is continuous, then this function f$f$ will be continuous also, and can be made total by linearly extending.
More generally, if h$h$ is bijection of [a,b]$[a,b]$ with another interval [a',b']$[a',b']$ disjoint from [a,b]$[a,b]$, then let f(x) = h(x)$f(x) = h(x)$ for x$x$ in [a,b]$[a,b]$, and f(x) = g(h-1(x))$f(x) = g\big(h^{-1}(x)\big)$ for x$x$ in [a',b']$[a',b']$. It follows that f(f(x)) = g(x)$f(f(x)) = g(x)$. And since there are 2|R|$2^{|\Bbb R|}$ many such functions h$h$, there are similarly many functions f$f$ satisfying the equation.
