Let $d \ge 2$, and consider the sphere $S^{d-1}$ embedded in $\mathbb R^d$. Does there exist a family of rotations $\{\mathcal O_v\}_{v \in S^{d-1}}$ which satisfies:

$\mathcal O_v e_1 = v$, and

$\mathcal O_v w = \mathcal O_w v$

for all $v, w \in S^{d-1}$ ? That is, the map $\mathcal O_v$ rotates the vector $e_1$ to $v$, and the action of rotating $w$ by $v$ is the same as rotating $v$ by $w$. The second property is a weak form of commutativity, as it is equivalent to $\mathcal O_v \mathcal O_w e_1 = \mathcal O_w \mathcal O_v e_1$.

If not, can we do it if we exclude a set of measure zero (such as the point $-e_1$)? That is, there exists a set $X \subseteq S^{d-1}$ of full measure such that above hold for all $v, w \in X$.

The case of $d=2$ is yes, for a very simple reason: $S^1 = \operatorname{SO}(2)$ is an abelian group, so we can define $\mathcal O_v$ as multiplication by $\operatorname{arg}(v)$, and the second property follows immediately by commutativity.

Here's what I am thinking for $d > 2$. Consider $S^{d-1}$ embedded as a subspace of $\mathbb R^d$. Let $v \ne -e_1$, so that there is a unique minimizing geodesic $\gamma_v$ from $e_1$ to $v$. Define $\mathcal O_v e_1 = v$. The vectors $e_2, \dots, e_d$ span the tangent space $T_{e_1}S^{d-1}$ at $e_1$, so define $\mathcal O_v e_i$ to be the result of parallel translation of $e_i$ along $\gamma_v$, for $i \ne 1$. Does this seem reasonable, or is it obviously false and I'm missing something?