The map
$$\psi_\alpha(z) = \frac{\alpha - z}{1 - \overline{\alpha}z}$$
can be shown to be a conformal map from the disc onto itself that interchanges $\alpha$ and $0$. I understand how to prove this in a few ways, although I'm a little bewildered at how one would guess this function $\psi_a$ in the first place. If I wanted an automorphism of the disc that interchanged $\alpha$ and $0$, how would I have thought of the above function?
Start from the fact that every automorphism of the Riemann sphere has the form $f(z) = \frac{az + b}{cz + d}$ (we'll tack on the requirement that $f$ sends the disc to itself later). To send $0$ to $\alpha$ we need $\frac{b}{d} = \alpha$ so we might as well set $b = \alpha, d = 1$. To send $\alpha$ to $0$ we need $\frac{a \alpha + \alpha}{c \alpha + 1} = 0$, so $a = -1$ and $c \neq \frac{1}{\alpha}$. So far we know that we must have $$f(z) = \frac{\alpha - z}{1 + cz}$$
for some $c$. Now we want this map to restrict to an automorphism of the open disc. This implies that $f$ must in fact preserve the unit circle: if $|z| = 1$ but $|f(z)| > 1$ then by moving $z$ into the open disc we get that $f$ doesn't send the open disc to itself, and if $|f(z)| < 1$ then by a similar argument we get that $f^{-1}$ doesn't send the open disc to itself.
This is a strong condition: we now know that $|z| = 1$ implies $|f(z)| = 1$. When $|z| = 1$ we have that $\bar{z} = \frac{1}{z}$, so setting $|f(z)|^2 = 1$ (remembering that $|w|^2 = w \bar{w}$) we conclude that $$\left( \alpha - z \right) \left( \bar{\alpha} - \frac{1}{z} \right) = |\alpha|^2 - \bar{\alpha} z - \frac{\alpha}{z} + 1 = \left( 1 + cz \right) \left( 1 + \frac{\bar{c}}{z} \right) = 1 + cz + \frac{\bar{c}}{z} + |c|^2$$
and equating coefficients of $z$ gives $c = - \bar{\alpha}$ as desired.
