 Solution

Write a Möbius transformation as $g(z)=(az+b)/(cz+d)$, with $ad-bc\ne 0$, on the Riemann sphere. If $c\ne 0$, infinity is not fixed, and its finite fixed points solveThe pole $-d/c$ cannot solve this equation, since substitution gives $(ad-bc)/c\ne 0$. The quadratic has one or two distinct complex roots, so these give exactly one or two fixed points.

If $c=0$, then $a,d\ne 0$ and infinity is fixed. If $a\ne d$, there is one additional finite fixed point, namely $b/(d-a)$. If $a=d$ and $b\ne 0$, the transformation is a nonzero translation, with no finite fixed point. If $a=d$ and $b=0$, it is the identity and fixes the entire sphere.

Thus a nonidentity Möbius transformation has one or two fixed points; the identity has infinitely many. The examples $z\mapsto z+1$, $z\mapsto 2z$, and $z\mapsto z$ realize the three possibilities respectively.