= Solution
Project from the north pole $N=(0,0,1)$ along the line through a sphere point onto the equatorial plane $Z=0$, identified with $\mathbb C$. Parameterizing this line gives
$$
\boxed{\pi(X,Y,Z)=\frac{X+iY}{1-Z},\qquad\pi(N)=\infty}.
$$
The inverse <stereographic projection> is
$$
(X,Y,Z)=\frac{(2\operatorname{Re}z,2\operatorname{Im}z,|z|^2-1)}{1+|z|^2}.
$$
Negating the three sphere coordinates consequently sends $z$ to $-1/\bar z$, with $0,\infty$ interchanged. This proves the <antipodal stereographic coordinate relation>.
A nonidentity <Möbius transformation> $T(z)=(az+b)/(cz+d)$ has $ad-bc\ne0$. For $c\ne0$, its fixed points are the roots of $cz^2+(d-a)z-b=0$, giving one or two distinct points. If $c=0$ and $a\ne d$, there is one finite root and the fixed point infinity. If $c=0$, $a=d$, and $b\ne0$, it is a translation and only infinity is fixed. The remaining case is the identity, already excluded. Thus the claim includes fixed points at infinity, not just finite solutions of the quadratic.
A nontrivial rotation of the sphere fixes exactly the two poles on its axis, which are antipodal. It therefore gives exactly two fixed stereographic points with $z_2=-1/\bar z_1$. Here the nonzero rotation angle is understood modulo $2\pi$; an angle that is a whole multiple of $2\pi$ would be the identity.
Conversely, use a sphere rotation whose Möbius map $S$ sends the two antipodal fixed points to $0,\infty$. The conjugate $STS^{-1}$ fixes both, hence equals $z\mapsto az$ for some $a\ne0$. If $|a|=1$, write $a=e^{i\theta}$; the inverse stereographic formula shows that this rotates $X+iY$ while leaving $Z$ fixed. Conjugating back by the rotation $S$ gives a sphere rotation. If $|a|<1$, $a^nz\to0$ for every finite $z$. If $|a|>1$, $a^nz\to\infty$ for every $z\ne0$, in the sphere topology. Pulling these conclusions back proves the <antipodal fixed-point classification of Möbius transformations>: \b[either the map is a sphere rotation or one fixed point attracts every point except the other fixed point].
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