I am interested in showing that isometries in $\mathbb{D}$ are either conformal self-maps in $\mathbb{D}$ or they are compositions of conformal self-maps with $z\mapsto \bar{z}$.
It is given that conformal self-maps in $\mathbb{D}$ are isometries. I think, to finish this, I just need to show that $z\mapsto \bar{z}$ is an isometry.
So far, I have considered the fact that the real interval $(-1,1)\subset \mathbb{D}$ is a geodesic in $\mathbb{D}$ that is fixed under $z\mapsto \bar{z}$. This means we are free to consider two points, find the geodesic that goes through them (say $\gamma$), map $\gamma$ conformally to $(-1,1)$, perform $z\mapsto \bar{z}$, and there will be no change in two initial points along $\gamma$ throughout this process. I mean, if $\rho(z_1,z_2)$ represents hyperbolic distance between $z_1$ and $z_2$ then we are allowed, by the previous argument, to say $\rho\left(\overline{\phi(z_1)},\overline{\phi(z_2)}\right)=\rho(z_1,z_2)$. But... this isn't quite what I want. What I want to show is that $\rho\left(\overline{z_1},\overline{z_2}\right)=\rho(z_1,z_2)$. Any suggestions? Thanks guys.
It is straightforward to verify that the curve of shortest hyperbolic length connecting $0$ to $s > 0$ is the line segment $[0,s]$. One computes
$$\rho(0,s) = \int_0^s \frac{2}{1-t^2}\,dt = \log \frac{1+s}{1-s}.$$
Since holomorphic automorphisms of the unit disk leave the hyperbolic length invariant, given two distinct $z,w \in \mathbb{D}$ we find
$$\rho(z,w) = \rho\left(0,\frac{w-z}{1-\overline{z}w}\right) = \rho\left(0,\left\lvert\frac{w-z}{1-\overline{z}w}\right\rvert\right) = \log \frac{1+\left\lvert\frac{w-z}{1-\overline{z}w}\right\rvert}{1-\left\lvert\frac{w-z}{1-\overline{z}w}\right\rvert}.$$
With this explicit formula for the hyperbolic distance, it is easily seen that the conjugation is an isometry.