 Solution

TakeIts 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, $M^{\prime }(1)=8/(3-1{)}^2=2$. 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 $|M^{\prime }(p)|$, 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.