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?)

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It may not be obvious at first, but this question seems to be related to [associative division algebras](https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29) (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](https://en.wikipedia.org/wiki/Quaternion#Matrix_representations):
$$
R\_1 = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\qquad
R\_2 = \begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}\qquad\\
R\_3 = \begin{pmatrix}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{pmatrix}\qquad
R\_4 = \begin{pmatrix}0&0&0&1\\0&0&-1&0\\0&1&0&0\\-1&0&0&0\end{pmatrix}\qquad
$$

Is this possible for any other $d$? I.e., can we find $R\_i$s even when there is no [associative division algebra](https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29) of dimension $d$?