Resolution of equation such that $f(...f(x)...)=x$ 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.
 A: 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?
