A disk self-map fixing $0$ cannot expand distance from $0$; equality forces a rotation.Suggest a slogan
Let \(f\colon\DD\to\DD\) be holomorphic with \(f(0)=0\). Then
\(\abs{f(z)}\leq\abs{z}\) for all \(z\in\DD\), and
\(\abs{f'(0)}\leq1\).
If \(\abs{f(z_0)}=\abs{z_0}\) for some \(z_0\in\DD\setminus\{0\}\), or if \(\abs{f'(0)}=1\), then \(f\) is a rotation: there is \(\lambda\in\CC\) with \(\abs{\lambda}=1\) such that \(f(z)=\lambda z\) for all \(z\in\DD\).