Is there a quick way to see that $\text{Aut}(D)$, the group of conformal automorphisms, acts transitively on the unit disk $D$?
I know that one can equivalently consider the projective special linear group acting on the upper half plane, but I can't see a rapid way to obtain transitivity from this.
It's a good idea to check $\mathrm{Aut} D=PSU(1,1)$. Its lift $SU(1,1)$ is conjugate to $SL_2\mathbb{R}$ within $SL_2\mathbb{C}$ via the Cayley transform which maps the upper half plane $H$ to the unit disk $D$, and $PSL_2\mathbb{R}=\mathrm{Aut}H$.
Also a good idea to show elements of $SU(1,1)$ are of the form $\big[\begin{smallmatrix}\alpha & \overline{\beta} \\ \beta & \overline{\alpha}\end{smallmatrix}\big]$ with $|\alpha|^2-|\beta|^2=1$. Keep in mind that describing $\mathrm{Aut}D$ explicitly like this is useful for dealing with all sorts of problems, not just this one.
To show $\mathrm{Aut}D\curvearrowright D$ transitively it suffices to check it can map any point of $w\in D$ to $0$. By setting $\big[\begin{smallmatrix}\alpha & \overline{\beta} \\ \beta & \overline{\alpha}\end{smallmatrix}\big]^{-1}0=-\overline{\beta}/\alpha=w$ we can solve it with $\alpha=1$ and $\beta=-\overline{w}$, which gives a matrix that is in $SU(1,1)$ up to a real multiple. This yields $\phi_w(z)=\frac{z-w}{1-\overline{w}z}$.
Alternatively, we can show $\mathrm{Aut}H\curvearrowright H$ is transitive, which would imply $\mathrm{Aut}D\curvearrowright D$ is via the Cayley transform. In fact the Borel subgroup of $\mathrm{Aut}H=PSL_2\mathbb{R}$ of upper triangular matrices $\big[\begin{smallmatrix} a & b \\ 0 & a^{-1} \end{smallmatrix}\big]$ acts regularly, there is a unique affine transformations $a^2z+b$ that turns $i$ into a given $w\in H$.