The Cauchy–Riemann equations say that if $u : \mathbb{C} \rightarrow \mathbb{C}$ is holomorphic then, regarded as a linear transformation of $\mathbb{R}^2$, its derivative is either zero or, up to a scalar, a rotation in the special orthogonal group $\mathrm{SO}_2(\mathbb{R})$. Explicitly,
$$ \left( \begin{matrix} \displaystyle \frac{\partial u}{\partial x} &\displaystyle \frac{\partial u}{\partial y} \\ \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y} \end{matrix} \right)_{z} = \left( \begin{matrix} \displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y} \\ \displaystyle- \frac{\partial u}{\partial y} & \displaystyle \frac{\partial u}{\partial x} \end{matrix} \right)_z = c \left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{matrix} \right) $$
where $c = \bigl( \frac{\partial u}{\partial x} \bigr)^2 + \bigl( \frac{\partial u}{\partial y} \bigr)^2$ is the determinant.
Fix $n \in \mathbb{N}$. What, if anything, can be said about the class of differentiable functions $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ whose derivative at every point in $\mathbb{R}^n$ is either zero or a scalar multiple of a special orthogonal matrix in $\mathrm{SO}_n(\mathbb{R})$?
I am of course willing to add extra assumptions, such as continuity of the derivatives, or even infinite differentiability, or that $f$ is a diffeomorphism, if this gives a more satisfying answer to the question.
