I am wondering if it exits a way to find "easely" the solutions of an equation about a function $f$ such that $$f^{(n)}(x)=x$$
where $f^{(n)}$ is the n-th composition of $f$ itself.
Obviously the identity is a trivial solution, I'm asking for all solution depending on $n$.
For example I know that for $n=4$, if $$f(x)=\frac{1+x}{1-x}$$ then $f^{(4)}(x)=x$
Any hints would be helpfull, thank you in advance.
Here's a general answer that might help you.
Suppose we have a rational function $$\rho(x)=\frac{ax+b}{cx+d}$$ and we define the "coefficient matrix" of $\rho$ to be $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$ then the coefficient matrix of $\rho \circ \rho$ or $\rho^2$ is $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} ^2$$ This theorem is rather trivial, and I will leave the proof to you.
If you know about rotation matrices, then you should know that $$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix} ^n=\begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \\ \end{pmatrix}$$ Meaning that if we let $$\theta=\frac{2\pi}{n}$$ Then we have $$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix} ^n=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ And so, since $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ is the coefficient matrix of the identity function $x$, if we take $\rho$ to be the rational function defined by the coefficient matrix $$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{pmatrix}$$ Then $$\rho^n(x)=x$$ and so $\rho$ can produce a solution to any functional equation in the form you asked about (though I'm sure it cannot be the only solution to any of them).
Does this help?