Can you partition the sphere into orthonormal bases? I've been writing some linear algebra problems with colleagues, and the following question occurred to us:

Let $S^2$ denote the unit sphere in $\mathbb{R}^3$. Does there exist a
partition $S^2=\bigsqcup_\alpha B_\alpha$ such that each $B_\alpha$ is
an orthonormal basis for $\mathbb{R}^3$?

The analogous question for the unit circle in the plane is easily affirmative, but that approach doesn't seem to work here.
 A: Using the Axiom of Choice, yes you can.
To get such a partition, start by enumerating all the points of the sphere with order type $\mathfrak{c}$ (the least ordinal number with the same cardinality as the sphere): say $\langle p_\alpha :\, \alpha < \mathfrak{c} \rangle$ is such an enumeration. Now we define our partition elements one at a time, via a transfinite recursion of length $\mathfrak{c}$. At stage $\alpha$ of the recursion, suppose we've already selected,  at previous stages of the recursion, some bases $\{ B_\xi :\, \xi < \alpha \}$ that will be in our partition. Now consider the point $p_\alpha$. There are two possibilities: either we already put $p_\alpha$ into one of the $B_\xi$'s for some $\xi < \alpha$, or we didn't. In the first case, we do nothing at stage $\alpha$ of the recursion: formally, we could define $B_\alpha = \emptyset$ in this case. In the second case, we choose an orthonormal basis $B_\alpha$ that contains $p_\alpha$, and that is disjoint from everything we put into our partition at an earlier stage. This is possible there is a $\mathfrak{c}$-sized collection $\mathcal C$ of orthonormal bases, any two of which intersect only in $p_\alpha$ (just rotate a basis a bit around $p_\alpha$); because $\bigcup_{\xi < \alpha}B_\xi$ contains $\leq 3 \cdot |\alpha| < \mathfrak{c}$ points, one of the bases in $\mathcal C$ contains no points from $\bigcup_{\xi < \alpha}B_\xi$. We choose some such basis to be $B_\alpha$. In the end, $\{ B_\alpha :\, \alpha < \mathfrak{c} \} \setminus \{\emptyset\}$ is the desired partition.
