= Solution
The <Schwarz lemma> states that a <holomorphic function> $f:\mathbb D\to\mathbb D$ with $f(0)=0$ satisfies
$$
\boxed{|f(z)|\le|z|,\qquad |f'(0)|\le1}.
$$
If equality holds in the first inequality at any nonzero point, or in the derivative inequality, then $f(z)=\lambda z$ for a constant $|\lambda|=1$. Conversely those rotations give equality.
To prove it, the function $g(z)=f(z)/z$ extends holomorphically at zero with value $f'(0)$, by the power series of $f$. On $|z|=r<1$, $|g(z)|\le1/r$. The <maximum modulus principle> gives this same bound inside that circle. For each fixed $z$, let $r\uparrow1$ to conclude $|g(z)|\le1$. This proves both inequalities. Equality in either makes $g$ attain its maximum modulus at an interior point, so the <maximum modulus principle> makes $g$ a constant of modulus one. This proves the equality cases as well.
For the classification of <automorphisms of the unit disk>, choose $a\in\mathbb D$ and define
$$
T_a(z)=\frac{z-a}{1-\overline a z}.
$$
Direct expansion gives
$$
1-|T_a(z)|^2=\frac{(1-|a|^2)(1-|z|^2)}{|1-\overline a z|^2}>0,
\qquad T_a^{-1}(w)=\frac{w+a}{1+\overline a w}.
$$
Thus $T_a$ is a <biholomorphism> of the disk and sends $a$ to zero. If $F$ is any <conformal map> of the disk onto itself, take $a=F^{-1}(0)$ and put $h=F\circ T_a^{-1}$. Both $h$ and $h^{-1}$ fix zero. The <Schwarz lemma> gives $|h(z)|\le|z|$ and, applied to $h^{-1}$ at $h(z)$, $|z|\le|h(z)|$. Equality holds, so $h(z)=e^{i\theta}z$. Consequently all the holomorphic conformal automorphisms are precisely
$$
\boxed{F(z)=e^{i\theta}\frac{z-a}{1-\overline a z},\qquad |a|<1,\quad\theta\in\mathbb R}.
$$
Every displayed map is indeed a conformal automorphism by the verified inverse and a rotation. Here conformal maps have the usual holomorphic, orientation-preserving meaning; if orientation reversal is allowed, their complex conjugates give the antiholomorphic alternatives.
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