= Solution
Write a <Möbius transformation> as $g(z)=(az+b)/(cz+d)$, with $ad-bc\ne0$, on the <Riemann sphere>. If $c\ne0$, infinity is not fixed, and its finite <fixed points> solve
$$
cz^2+(d-a)z-b=0.
$$
The pole $-d/c$ cannot solve this equation, since substitution gives $(ad-bc)/c\ne0$. The quadratic has one or two distinct complex roots, so these give exactly one or two <fixed points>.
If $c=0$, then $a,d\ne0$ and infinity is fixed. If $a\ne d$, there is one additional finite <fixed point>, namely $b/(d-a)$. If $a=d$ and $b\ne0$, 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 \b[a nonidentity <Möbius transformation> has one or two <fixed points>; the identity has infinitely many]. The examples $z\mapsto z+1$, $z\mapsto2z$, and $z\mapsto z$ realize the three possibilities respectively.
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