Proof that automorphisms of unit disk are of the form $\lambda \dfrac{z-a}{\overline{a}z-1}$

Question

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?

Answer 1

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.