Which Möbius transformations send 0, 1 and infinity to 0, 1 and infinity 
Possible Duplicate:
How do I find out the symmetry of a function? 

Let $f:\mathbf{P}^1 \longrightarrow \mathbf{P}^1$ be a Möbius transformation $z\mapsto (az+b)/(cz+d)$ sending $\{0,1,\infty\}$ to $\{0,1,\infty\}$ with $ad-bc = 1$. 
I suspect there are only a finite number of such Möbius transformations. What are these?
A non-trivial example is $f(z) = 1/(1-z)$. It sends $0$ to $1$, $1$ to $\infty$ and $\infty$ to $0$. Note that  $f(z) = -z+1$ is not an example, because $ad-bc = -1$ in this case.
 A: If $0\mapsto 0$, then $b=0$. This means that $d=\frac{1}{a}$. 


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*If $f(1)=1$ and $f(\infty)=\infty$, then we must have $a=c+d$ and $c=0$, so $ad=1$, $a=1$. Hence $a=d=\pm 1$, $b=c=0$. The only transformation is the identity.

*If $f(1)=\infty$ and $f(\infty)=1$, then we must have $a=c$ and $c+d=0$; since $d=\frac{1}{a}= -c = -a$, we have $-a^2=1$, so $a=c=\pm i$, $d=\mp i$.
If $0\mapsto 1$, then $b=d$, so $(a-c)d=1$. 


*

*If $1\mapsto 0$ and $\infty\mapsto\infty$, then $a=-b$ to get $1\mapsto 0$, and $c=0$ to get $\infty\mapsto \infty$. Then $(a-c)d = -d^2=1$, so $b=d=\pm i$, $a=\mp i$. (corrected; apologies to the anonymous user who suggested the edit)

*If $1\mapsto\infty$ and $\infty\mapsto 0$ then we must have $a=0$, and $c+d=0$. So $1 = (a-c)d = -cd = d^2$. Hence $d=b=\pm 1$, $c=\mp 1$, $a=0$. 


If $0\mapsto\infty$, then $d=0$, so $bc=-1$. 


*

*If $1\mapsto 0$ and $\infty\mapsto 1$, then $a=c$ and $a+b=0$. So $-1 = bc = -ac = -c^2$, hence $a=c=\pm 1$, $b=\mp 1$.

*If $1\mapsto 1$ and $\infty\mapsto 0$, then $a=0$ and $b=c$. Hence $-1=bc=b^2$, so $b=c=\pm i$, $a=d=0$. 

