Consider two points on the unit sphere, $C_1$ and $C_2$. Let these points be close enough such that circles of radius $r$ drawn around each point intersect at two points, $N_1$ and $N_2$. The vectors that point to $N_1$ and $N_2$ can be found by the following formula:
$$ f(C_1, C_2) = k\frac{\vec{C_1}+\vec{C_2}}{|\vec{C_1}+\vec{C_2}|} \pm \sqrt{1-k^2}\frac{\vec{C_1}\times\vec{C_2}}{|\vec{C_1}\times\vec{C_2}|} $$
Where $k = \cos(\theta)/\cos(\alpha/2)$, $\theta$ is the half angle of the circle from the origin and $\alpha$ the angle between the vectors pointing to $C_1$ and $C_2$.
If one interprets the $\pm$ as producing two separate vectors, then $f(\vec{C_1},\vec{C_2}) = (\vec{N_1},\vec{N_2})$. Furthermore, the function is almost it's own inverse, that is: $f(\vec{N_1},\vec{N_2}) = (\vec{C_2},\vec{C_1})$. I say almost since the order of $C_1$ and $C_2$ is reversed. It takes four applications of $f$ to return to the original configuration:
$$ f^{(4)}(\vec{C_1},\vec{C_2}) = (\vec{C_1},\vec{C_2}) $$
Question:
Why $f^{(4)}(x) = x$? I was surprised that I couldn't simply make this function $f^{(2)}(x) = x$. Is there a good reason for this on $\mathbb{R}^3$? Is this sort of behavior categorizable? Perhaps as a mini-lecture on group theory? An accepted answer would explain if $f^{(2)}(x) = x$ is possible in this context, and if not explain why.