The biholomorphisms \(\DD\to\DD\) are exactly the maps \[ f(z)=\lambda\,\frac{\alpha-z}{1-\bar\alpha z},\qquad \alpha\in\DD,\ \lambda\in\CC,\ \abs{\lambda}=1. \]
Write \(\psi_\alpha(z)\coloneqq\frac{\alpha-z}{1-\bar\alpha z}\) for the Blaschke factor at \(\alpha\in\DD\); it is a biholomorphism \(\DD\to\DD\) with \(\psi_\alpha\circ\psi_\alpha=\operatorname{id}\), so \(\lambda\psi_\alpha\) is a biholomorphism of \(\DD\) for \(\abs{\lambda}=1\).
Conversely, let \(f\colon\DD\to\DD\) be a biholomorphism, let \(\alpha\coloneqq f^{-1}(0)\in\DD\), and put \(g\coloneqq f\circ\psi_\alpha\). Then \(g\) is a biholomorphism of \(\DD\) with \(g(0)=f(\alpha)=0\). The Schwarz lemma applied to \(g\) and to \(g^{-1}\) gives \(\abs{g'(0)}\le1\) and \(\abs{(g^{-1})'(0)}=1/\abs{g'(0)}\le1\), so \(\abs{g'(0)}=1\) and \(g(z)=\lambda z\) with \(\abs{\lambda}=1\). Hence \(f=g\circ\psi_\alpha=\lambda\psi_\alpha\).