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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 givesThe inverse stereographic projection isNegating 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 $ad-bc\ne 0$. For $c\ne 0$, its fixed points are the roots of $c{z}^2+(d-a)z-b=0$, 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 $b\ne 0$, 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 $ST{S}^{-1}$ fixes both, hence equals $z\mapsto az$ for some $a\ne 0$. 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 $|a|<1$, $a^{n}z\to 0$ for every finite $z$. If $|a|>1$, $a^{n}z\to \infty$ for every $z\ne 0$, in the sphere topology. Pulling these conclusions back proves the antipodal fixed-point classification of Möbius transformations: either the map is a sphere rotation or one fixed point attracts every point except the other fixed point.