 Solution

True. Conjugate the unique fixed point to infinity. The resulting map lies in the Affine subgroup of the Möbius group, so it has the form $g(z)=az+b$. A unique fixed point forces $a=1$ and $b\ne 0$; otherwise $g$ would have a finite fixed point as well. Conjugation by the scaling $z\mapsto z/b$ changes $z\mapsto z+b$ into $z\mapsto z+1$. Thus a nonidentity Möbius transformation with one fixed point is conjugate to a translation.