Suppose $G$ is a subgroup of $\text{Aut}(D(0,1))$ which contains all the origin -fixing rotations and at least one element other than these. Then the problem asks me to prove that $G$ is all of $\text{Aut}(D(0,1))$.
Firstly ,$G$ contains an element of the form $\phi_{\alpha,\theta}(z)=e^{i\theta}\frac{\alpha-z}{1-\overline{\alpha}z}$ where $\alpha$ is a non-zero element of the open unit disc. Since all $e^{i\psi}z$ are inside $G$ , all $\phi_{\beta,\psi}(z)=e^{i\psi}\frac{\beta-z}{1-\overline{\beta}z}$ (where $|\beta|=|\alpha|$ and $\psi\in[0,2\pi)$) are in $G$.Now if $c>0$ with $c=|\alpha|$ then $ \phi_{c,\theta}\circ\phi_{c,\theta}(z)=\phi_{c\frac{1+e^{i\theta}}{e^{i\theta}+c^2},\psi}$ for some $\psi$. Now varying $\theta$ and using the reasoning of the previous line we get all elements $\phi_{\alpha,\theta}(z)$ where $0\leq|\alpha|\leq\frac{2c}{1+c^2}$ are in $G$. To finally get all pairs $\alpha, \theta$ we can consider $ \phi_{\frac{2c}{1+c^2},\theta}\circ\phi_{\frac{2c}{1+c^2},\theta}$ and keep on repeating the process with the largest real $\alpha$ we get at each stage . But I'm stuck on how to show that in this way we obtain all $\alpha$.
Can someone please help me on this? thanks.
Let $\phi_0=\phi_{\alpha,\theta}(z)=e^{i\theta}\frac{\alpha-z}{1-\overline{\alpha}z}$ the given element of $G$; note that $\phi_1=e^{-i\theta}\phi_0 \in G$ and also if $\arg \alpha=\theta_0, |\alpha|=r_0$ then $\phi_2(z)=\phi_1(-e^{i\theta_0}z)=\frac{r_0+z}{1+r_0z} \in G$ and by the same procedure one has $\frac{\alpha_1-z}{1-\overline{\alpha_1}z} \in G$ for all $|\alpha_1|=r_0$
Also note that if $|w|=r_1, |y|=r_2$ and there is $\phi \in G, \phi(w)=y$ then for every $|w_1|=|w|, |y_1|=|y|$ there is $\phi_{y_1,w_1} \in G, \phi_{y_1,w_1}(w_1)=y_1$ by rotating $\phi$ both inside and outside.
We know that the disc automorphism group is transitive and moreover that if $|w| \ne 0$ and $\psi_1(0)=\psi_2(0)=w$, then $\psi_1^{-1} \circ \psi_2$ is a rotation (hence in $G$) by Schwarz Lemma, so it is enough to show that for every $w$ there is $\psi \in G, \psi(0)=w$, while the paragraph above shows that it is enough to show that for every $r<1$ there is a $\psi_r \in G, |\psi_r(0)|=r$
But now $0 \le |\frac{\alpha_1-r_0}{1-\overline{\alpha_1}r_0}| \le \frac{2r_0}{1+r_0^2}=c_0$ when $\alpha_1$ varies on the circle of radius $r_0$ with equality at both ends for $\alpha_1= \pm r_0$ so by intermediate value theorem $|\phi_{\alpha_1}(r_0)|$ takes all values in $[0,c_0]$ which by the above means that for all $0 \le r \le c_0$ there is $\eta_r \in G, |\eta_r(r_0)|=c_0$ and since $\eta_r\circ \phi_r(0)=\eta_r(r_0)$, we get $\psi_r \in G, |\psi_r(0)|=r, 0 \le r \le c_0$ which is almost what we need except that we need $c \to 1$
But now looking at $\phi_n=\phi_2 \circ \phi_2 \circ ...\circ \phi_2$ with composition taken $n$ times, it is easy to see that $\phi_n(0) \to 1$ (increasing sequence of bounded positive numbers satisfying $x_{n+1}=\frac{2x_n}{1+x_n^2}$ so they have a finite positive limit $l=\frac{2l}{1+l^2}$ so $l=1$) , which combined to the fact that if $r \to 1$, then $c_r=\frac{2r}{1+r^2} \to 1$ and the above considerations, solves the problem!