Composition of Möbius Transformation to Identity Let $f(x) = \frac{ax+b}{cx+d}$ with $x \in \mathbb{R} \setminus \left \{ -\frac{d}{c} \right \} $ be the Möbius Transformation. Find all $a,b,c,d$ such that
$$ \underbrace{f \circ f \circ \dots \circ f}_{n \text{ times}} = \text{id}.$$
My professor and I came to the conclusion that you probably can't write down nice forms depending on $n$. (The case $n=3$ gives a complicated solution.)
However, can you think of any "good" condition or any meaningful progress towards this problem? Feel free to change it a little - it should just be approximately this problem! 
(Note that it's easy to find a solution $a,b,c,d$ that works for all $n$. The problem is to find all solutions.)
 A: NOTATION. We denote $f^{\circ n}:=f\circ f\circ \ldots \circ f$.
I cannot find a "pointwise" condition on $a,b,c,d$. (EDIT: see bottom of the post for such a condition, due to Will Sawin). However if we regard the matrix 
$$
A=\begin{bmatrix} a & b \\ c & d\end{bmatrix},$$ 
then interpreting the Möbius transformation as a projective transformation we see that the given Möbius transformation sastisfies $f^{\circ n}=\mathrm{id}$ if and only if $A^n$ is a scalar multiple of $I=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$. As shown here such matrices are necessarily diagonalizable, and the equation 
$$
A^n=zI$$
shows that the eigenvalues $\lambda_1, \lambda_2$ of $A$ must satisfy 
$$\lambda_1^n=\lambda_2^n=z.$$

Conclusion. The condition $f^{\circ n} =\mathrm{id}$ for the Möbius transformation represented by the invertible matrix $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is equivalent to $A$ being diagonalizable and the eigenvalues $\lambda_1, \lambda_2$ of $A$ satisfying $\lambda_1^n=\lambda_2^n$.

EDIT. A necessary condition for $f^{\circ n}=\mathrm{id}$ is the following: 
$$\tag{1}
\frac{(a+d)^2}{ad-bc}=2(1+\cos\theta),\quad \theta=\frac{2\pi}{n}k,\ k\in\mathbb Z.$$ 
If $k\ne 0 \mod n$, the condition is also sufficient. Otherwise, it is not, and the necessary and sufficient condition is that (1) holds and $A$ is diagonalizable. 
Proof. This is due to Will Sawin. Letting $\lambda_1, \lambda_2$ denote the eigenvalues of $A$, we have 
$$ 
\frac{(a+d)^2}{ad-bc} = \frac{\lambda_1}{\lambda_2}+\frac{\lambda_2}{\lambda_1}+2,$$ 
which equals $2+2\cos\theta$ precisely when $\lambda_1/\lambda_2 = e^{2\pi i \frac k n}$, that is, $\lambda_1^n=\lambda_2^n$. If $k\ne 0 \mod n$, then $\lambda_1\ne \lambda_2$ and this automatically implies that $A$ is diagonalizable. Therefore $f^{\circ n}=\mathrm{id}$ because of the characterization proven above.
On the other hand, the matrix $A=\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$ satisfies (1) with $k=0$ and $$A^n= \begin{bmatrix} 1 & n\\ 0 & 1\end{bmatrix},$$ so it cannot satisfy $A^n=A$ for any $n>1$. Notice that $A$ is not diagonalizable.
To conclude, we note that in the case $k\ne 0\mod n$, we need to manually add to (1) the assumption that $A$ is diagonalizable to ensure that $f^{\circ n}=\mathrm{id}$ holds. $\square$
