= Solution
Elements $g,h\in G$ are conjugate when
$$
h=xgx^{-1}
$$
for some $x\in G$. Then
$$
h^n=xg^nx^{-1}
$$
for every integer $n$, so $h^n=e$ exactly when $g^n=e$. They therefore have the same <order of a group element>[order].
The Möbius group consists of the <Möbius transformation>[Möbius transformations]
$$
z\longmapsto\frac{az+b}{cz+d},
\qquad ad-bc\ne0,
$$
of the Riemann sphere, under composition. If $h=xgx^{-1}$, then
$$
h(x(p))=x(g(p)).
$$
Thus $p$ is fixed by $g$ exactly when $x(p)$ is fixed by $h$. Conjugation gives a bijection between their fixed-point sets, so conjugate elements have the same number of fixed points.
A nonidentity Möbius transformation has at most two fixed points because its fixed-point equation is quadratic on the Riemann sphere. If it has only one repeated fixed point, conjugate that point to infinity; the transformation becomes a nontrivial translation $z\mapsto z+c$, which has infinite order. Consequently every nontrivial finite-order element has two distinct fixed points:
$$
\boxed{|\operatorname{Fix}(g)|=2}.
$$
This is <fixed points of a finite-order Möbius transformation>.
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