Finding $f$ such that $f(f(f(f...(x)))) = x$ I would like to find the set of continuous functions $f_n(x)$, where $f_n(x):\mathbb{R}\to \mathbb{R}$ satisfies $$f_n(f_n(f_n(f_n...(x)))) = x$$ where there are $n$ iterations of $f(x)$. For example $f_1(x)$ would be the solution to $f_1(x)=x$. $f_2(x)$ would be the solution to $f_2(f_2(x)) = x$.
For $f_1(x)$, the only solution is $f_1(x)=x$. For $f_2(x)$, the solutions are involutions.
For $f_3(x)$, the only answer is $f_3(x)=x$.
For all other $f_n(x)$, one solution is $f_n(x) = x$.
My question: For $n \ge 3$, is $f_n(x) = x$ the only solution? If not, what are the solutions?
Edit: @MattSamuel said that any involution works for an even $n$. This is because $f_n(f_n(x))$ can be replaced with $x$. For example,  $$f_2(f_2(f_2(f_2(f_2(f_2(x)))))) = f_2(f_2(f_2(f_2(x)))) = f_2(f_2(x)) = x$$ However, this does not necessarily mean that involutions are the only set of solutions for $f_{2k}(x)$.
 A: There are no monotone solution for odd $n$ other then $f_n(x) = x$. There are also no continuous solutions as any continuous injection is monotone.
There are many discontinuous solutions for any $n$. Represent any real number $x$ as $x = \lfloor x \rfloor + \{x\}$ (integer and floor part), so we now have bijection $\mathbb{R} \leftrightarrow [0, 1) \times \mathbb{Z}$. Choose your favorite bijection $g_n: \mathbb Z \leftrightarrow \mathbb Z$ of order $n$ - for example, $g_n(i) = (i + 1) \mod n + \lfloor\frac i n\rfloor$ - split $\mathbb{Z}$ into segments of length $n$ and rotate any segment. Now define $f_n(x) = \{x\} + g_n(\lfloor x \rfloor)$. It's even continuous everywhere but in points $n - 1 + kn$.
(for simplicity, I'll denote $f^n$ to be $n$-th iteration of $f$ - we will not need powers here)
The only continuous solutions are involutions - answer you linked can be extended to proof it. $f$ have to be monotonic - if it isn't - say we have $f(x) > f(y) > f(z)$ while $x > z > y$ - then it's not injective, as there is point in $q \in [z, x]$ s.t. $f(q) = f(y)$, so we will have $f^n(q) = f^n(y)$ but $q \neq y$.
If $f$ is strictly increasing, then $f(x) = x$ by @Najib's argument.
If $f$ is strictly decreasing, then $f$ has a single fixed point $x_0$. We have $f(x_0 + a) < f(x_0) = x_0$ and $f(x_0 - a) > f(x_0) = x_0$ for positive $a$.
$g =f\circ f$ is continuous and injective - so monotonic. As $f(x_0 + 1) < f(x_0)$, we have $f(f(x_0 + 1)) > x_0$, so $g$ is increasing. If $g(x) \neq x$ for some $x$, we will again have $f^n(x) \neq x$. So $g(x) = x$. And thus $f$ is involution.
