I came up with this question while trying to solve an exercise:
Let $f: \{z:|z|<1\} \to \{z:|z|<1\}$ be analytic, and $f(0)=0$ - the Schwarz lemma case. From the lemma it follows that $$\forall 0<r<1: f(\{z:|z|<r\})\subseteq \{z:|z|<r\}$$
If there exists some specific $0<r<1$ such that $f(\{z:|z|<r\}) = \{z:|z|<r\}$, does it follow that $f(z)=cz$ for some $|c|=1$? If not, is $f$ of some simple form?
Let $0<r<1$ be the radius that satisfies $f(\{z:|z|<r\}) = \{z:|z|<r\}$.
Let $w_n:=r-\frac{1}{n}$.
Since $f(\{z:|z|<r\}) = \{z:|z|<r\}$, for every $w_n$ there exists a $z_n \in \{z:|z|<r\}$ such that $f(z_n)=w_n$.
$\{z_n\}_{n=1}^{\infty}$ is a bounded sequence and therefore contains a convergent subsequence $\{z_{n_k}\}_{k=1}^{\infty}$, that converges to some $|z_0|\leq r$. Moreover, $|z_0|=r$, otherwise $f$ would have been constant on $\{z:|z|<r\}$ by the maximum principle.
Since $f$ is continuous,
$\lim_{k\to\infty} z_{n_k}=z_0 \Longrightarrow \lim_{k\to\infty} f(z_{n_k})=f(z_0)$,
and $f(z_{n_k})=w_{n_k}\xrightarrow{k \to \infty}r$, so $f(z_0)=r$, which means $|f(z_0)|=|z_0|$.
Since $0<|z_0|<1$, the Schwarz lemma gives that $f(z)=cz$ for some $|c|=1$.