Lets denote the upper half-plane by $\mathbb{H}$ and the ball centered at origin as $\mathbb{D}$. My question is what is wrong with my reasoning because I think something should be wrong but I can't catch what it is. Let me write it down.
Automorphisms of upper half plane are given by
$$h(z)=\frac{az+b}{cz+d} \ \ \ \text{where} \ \ \ ad-bc>0$$ These are called bilinear transformations (with $ad-bc\neq 0$) in general.
We also know from Bak & Newman Proposition 13.19 that
- A bilinear transformations (other than the identity mapping $f (z) = z$) has at most two fixed points.
I've also proved that if an analytic map from $\mathbb{D}$ to $\mathbb{D}$ has 2 distinct fixed points then it is identity. ( Chapter 8 Exercise 12 Stein-Shakarchi Complex-Analysis )
Here is where the problem begins. We can map $\mathbb{H}$ to $\mathbb{D}$ using the conformal map $$ F(z)=\frac{i-z}{i+z}$$ Now suppose $G$ is an automorphism (conformal map with same domain and range) that fixes 2 points, $z_1,z_2$, on $\mathbb{H}$. Using $F$ we can construct a new automorphism $F\circ G\circ F^{-1} :\mathbb{D}\to \mathbb{D}$. This again fixes 3 points, since $$(F\circ G\circ F^{-1})(F(z_{1,2}))=F(z_{1,2}) $$ But by the exercise I proved, the map $F\circ G\circ F^{-1}$ is identity. This means \begin{align} &F\circ G\circ F^{-1}(z)=z \\ \implies &F\circ G(z)= \frac{i-z}{i+z} \\ \implies &\frac{i-G(z)}{i+G(z)}= \frac{i-z}{i+z} \end{align} Which means $G(z)=z$. But if this is the case, why wouldn't they just say a bilinear transformation has at most one fixed point.