For which integers $d \geq 1$ can we find real matrices $R_1, \dotsc, R_d$ of size $d \times d$ such that for any unit vector $v \in \mathbb{R}^d$, $$R_1 v, \dotsc, R_d v$$ is an orthonormal basis? Note that the chosen set of $R_i$s has to work simultaneously **for all** $v$. (Does this phenomenon have a name?)

It may not be obvious at first, but this question seems to be related to associative division algebras (i.e., reals, complex numbers, and quaternions). Indeed, I know that such matrices $R_i$ can be found when $d=1,2,4$:

- $d=1$ is trivial—simply take $$R_1 = \begin{pmatrix}1\end{pmatrix}$$
- $d=2$ is based on
*complex numbers*—we represent $1$ and $i$ by $2 \times 2$ real matrices as follows: $$ R_1 = \begin{pmatrix}1&0\\0&1\end{pmatrix}\qquad R_2 = \begin{pmatrix}0&-1\\1&0\end{pmatrix} $$ - $d=4$ is based on
*quaternions*—we represent $1$, $i$, $j$, $k$ by $4 \times 4$ matrices as follows: $$ R_1 = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\qquad R_2 = \begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}\qquad\\ R_3 = \begin{pmatrix}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{pmatrix}\qquad R_4 = \begin{pmatrix}0&0&0&1\\0&0&-1&0\\0&1&0&0\\-1&0&0&0\end{pmatrix}\qquad $$

Is this possible for any other $d$? I.e., can we find $R_i$s even when there is no associative division algebra of dimension $d$?