$S^{n}$ be the unit sphere in $\mathbb{R}^{n+1}$. Let $\pi_N$ be stereographic projection from the sphere without the north pole on to $\mathbb{R}^{n}$ and let $\pi_S$ be defined similarly using the south pole.
On $\pi_N(\pi_S^{-1}(\mathbb{R^n}))=\mathbb{R^n-0}$, one can take $\pi_S \circ \pi_N^{-1}$. By messy algebra, I showed that this is the inversion through the unit sphere $S^{n-1} \subset \mathbb R^{n}$ sending $x \in \mathbb{R}^{n} \mapsto x/|x|^2$.
In full detail, I showed that that under this correspondence $z \mapsto z \cdot 2/(1+|z|^2)+(|z|^2-1)/(|z|^2+1)(0,....,0,1) \mapsto z \cdot (2/(1+|z|^2))/(2 |z|^2 /(1+|z|^2))=z/|z|^2$
Is there a geometric way of seeing this using just inner products and such. i.e. I don't want to have to use the quadratic forumula like I did (I am trying to do this along the lines of the symmetry lemma given in Axler's Harmonic function theory. You don't have to though.)
The points, its projections and the poles all reside in some 2D cross section of space, so it suffices to do the problem in 2D. I assume you're working with the unit circle.
Let's draw a picture and label some points. Say $N$ is the north pole, $S$ is the south pole, $O$ is the origin, $Q$ is a point on the unit circle (without loss of generality in quadrant IV), with stereographic projections given by $\pi_S(Q)=P$ and $\pi_N(Q)=R$.
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We will show the triangles $\triangle NOR$ and $\triangle POS$ are similar by showing they have the same angle measures (in order). Similarity implies an equality of ratios $[PO:OS]=[NO:OR]$. Since both of the line segments $NO$ and $OS$ are length $1$, this proves $PO$ and $OR$ are reciprocal lengths.
By the "angle inscribed in semicircle" theorem, $\angle NQS$ is a right angle. Clearly $\angle NOR$ is also a right angle. Note that the angles $\angle NRO$ and $\angle PRQ$ are equal, and complementary to $\angle ONR$ and $\angle OPS$ which are in turn complementary to $\angle NSP$.