Proof that automorphisms of unit disk are of the form $\lambda \dfrac{z-a}{\overline{a}z-1}$ I am trying to show that the most general Möbius map that sends the unit disk to the unit disk is $\lambda \dfrac{z-a}{\overline{a}z-1}$ where $|\lambda|=1$ and $|a|<1$. I have shown that the maps of this form are a subgroup of the set of Möbius maps, and that they are automorphisms of the unit disk. I want to show that there are no other maps which do this. The proof I have seen considers an arbitrary Möbius map $f(z)=\dfrac{az+b}{cz+d}$ and then composes this with the map $g(z)=\lambda\dfrac{z-f(0)}{\overline{f(0)}z-1}$, which supposedly takes us back since $g(f(0))=0$ and if we take an arbitrary point on the boundary of the unit circle we know that $f$ and $g$ will send it to the boundary, and then we can choose $\lambda$ to rotate it so that it goes back to its original position. So $g$ is the inverse of $f$ and since those maps form a group $f$ must be of that form too. I am not sure why considering two points is sufficient to know that $(g \circ f) (z)=z$ though. Don't we need three points to determine a Möbius map?

Conceptually, the third degree of freedom is already used by the requirement that the map be an automorphism of the unit disk, so only two points are needed to specify a Möbius automorphism of the unit disk. 
In terms of the proof itself, it may be easiest to apply the Schwarz lemma,  which states (in one of its forms) that a function which maps the unit disk holomorphically onto itself and satisfies $f(0)=0$ must be a rotation. 
