Isometry of $S^2$ taking a point to $N=(1,0,0)$ While reading a related proof, the author of my geometry textbook states that there exists an isometry of the $2-Sphere$ or $S^2$ that can take any arbitrary point to the north pole.
I know that there are three types of isometries of $S^2$ namely a reflection in a line (about two antipodal points), a reflection in a great circle, and a spin reflection, but I'm unsure which of these (or what combination thereof) could result in the desired isometry.
 A: Any rotation about a diameter of the sphere will provide an isometry. For the arbitrary point in the question, there is a diameter that is perpendicular to each of the radii that join the centre to the arbitrary point and to the point $(1,0,0)$.
In general, if $\theta$ is the angle between these two radii, then rotatating about the diameter described above by $\theta$ will take the point to $(1,0,0)$ (taking care to get the sense of rotation right).
If an explicit calculation of the rotation matrix is needed, then the method described in this question for rotating one vector onto another can be used.
A: I'm fairly sure any kind of isometry could be made to work. 
Given any point $P \neq N$, there's a great circle $C$ connecting $N$ and $P$ (it's unique, if $N$ and $P$ aren't antipodal). The points divide $C$ into two arcs, and the midpoints of these two arcs are antipodal. A half-turn about the line connecting these antipodal points would interchange $N$ and $P$ (there's also a rotation "along" $C$ that sends $P$ to $N$).
Alternatively, there's a great circle $C'$ perpendicular to $C$ and containing the two antipodal points above. A reflection through $C'$ also interchanges $N$ and $P$.

Since points of the $2$-sphere correspond to directions in $\Bbb R^3$, if you believe there are isometries of $\Bbb R^3$ that can send any $1$-dimensional subspace to the $z$-axis, this is (essentially) equivalent to what the author of the text is claiming. 
