Powers of Möbius transformations equal to identity? I'm looking at "Mobius transformations" where $a,b,c,d\in\mathbb R$. I want to know for which $n$ there exists $a,b,c,d$ such that for $f(x) = \dfrac{ax+b}{cx+d}$,
$$f^n(x) = f(f(...(f(x)))) = x$$
and what relationships between $a,b,c,d$ are required. Or if it is for all $n$, if there is a pattern to these relationships. 
For example,
$$f^1(x)=x \iff a-d=0, c= 0, b=0, a\neq 0$$
$$f^2(x)=x \iff a+d=0, a^2+bc\neq 0$$
I see that for $f^{2k}(x)$, we can get an iterative relationship from the above. With the same conditions as the $n=2$ case.

As said by @Did, you need to find coefficients a,b,c,d such that 
$$M(f)^n=\begin{pmatrix}a&b\\c&d\end{pmatrix}^n=k\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
Thinking to rotation matrices, there is an evident solution :
$$M=\begin{pmatrix}\cos(\frac{\pi}{n})&-\sin(\frac{\pi}{n})\\ \sin(\frac{\pi}{n})&\cos(\frac{\pi}{n})\end{pmatrix}$$
Otherwise said, a possible Möbius (or homographic) transformation is :
$$f_n(x)=\dfrac{\cos(\frac{\pi}{n})x-\sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})x+\cos(\frac{\pi}{n})}$$
Edit : This rotation is not unique in general. Let us take an example: if $n=12$, you can take any constant $K=1,2,\cdots 11$ in the following matrix 
$$M=\begin{pmatrix}\cos(\frac{K\pi}{n})&-\sin(\frac{K\pi}{n})\\ \sin(\frac{K\pi}{n})&\cos(\frac{K\pi}{n})\end{pmatrix}$$
and have $M^n=\pm I_2.$
(following a very judicious remark of "studiosus") a very general type of non trivial matrices $M$ such that $M^n=\pm I_2$, at least among diagonalizable matrices is obtained by thinking to the conjugation operation, that doesn't change the eigenvalues that will still be $e^{iK\pi/n}$ and $e^{-iK\pi/n}$:
$$M=P\begin{pmatrix}\cos(\frac{K\pi}{n})&-\sin(\frac{K\pi}{n})\\ \sin(\frac{K\pi}{n})&\cos(\frac{K\pi}{n})\end{pmatrix}P^{-1}$$
for any invertible matrix $P.$
