Since we have a conformal equivalence $\psi :\mathbb{H}\to \mathbb{D}$, given by $\frac{z-i}{z+i}$, by the previous theorem we have that $Aut(\mathbb{H})$ is given by 1;; all $f(z)=\frac{az+b}{cz+d}$ for $a,b,c,d$ real and $det(Mat(f))>0$.