= Solution
Take
$$
\boxed{M(z)=\frac{3z-1}{3-z}.}
$$
Its defining <matrix> has nonzero <determinant> eight, so it is a <Möbius transformation>. Direct substitution shows that it fixes $p=1$ and $q=-1$, which satisfy $p\overline q=-1$. However, $M'(1)=8/(3-1)^2=2$. Differentiating the inverse projection formula gives the round spherical line element $ds=2|dz|/(1+|z|^2)$. At a fixed finite point, its local length multiplier under a <Möbius transformation> is exactly $|M'(p)|$, since the metric factors in numerator and denominator agree. Here that multiplier is two, so this map is not an <isometry> and cannot be a <rotation in three dimensions> of the sphere. It demonstrates that <antipodal fixed points do not characterize sphere rotations>.
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