Given a function $u$, how to find an orthogonal matrix $A$ such that each component of $A \nabla u$ has same $L^2$ norm? Let $u$ be a differentiable function defined in a bounded domain $\Omega\subset \mathbb{R}^n$, with $\nabla u \in L^2(\Omega)$. Show that there is an orthogonal matrix $A$ such that for any $1 \le i \le n$, we have $$\Vert(row_iA) \cdot \nabla u\Vert_{L^2(\Omega)}=\frac{1}{\sqrt{n}}\Vert \nabla u \Vert_{L^2(\Omega)}.$$

In two dimensions, this can be easily proved by setting $A$ to be a rotation matrix with parameter $\theta$, which is solvable due to the above equation. In three dimensions, I cannot find an elegant argument to prove the above statement. Any ideas or comments are really appreciated.
 A: Up to notational differences, your question is summed up in the following:
Proposition. Let $E$ be an $n$-dimensional linear space with scalar product, $(\Omega,\mu)$ a measure space, and $V \colon \Omega \to E$ a function with $\int |V|^2 d\mu = 1$. Then there is an orthonormal basis $e_1,\ldots,e_n$ of $E$ for which
$$
\| V \cdot e_1 \|^2_{L^2} = \| V \cdot e_2 \|^2_{L^2} = \ldots = \| V \cdot e_n \|^2_{L^2} = \frac 1n.
$$
To see this, just take $V$ to be the gradient $\nabla u$ normalized by its $L^2$-norm, and $A$ to be the related change-of-basis matrix.
$\newcommand{\R}{\mathbb{R}}$
Proof. To rephrase the statement, let me introduce the operator
$$
T(e_1,\ldots,e_n) = \left( \| V \cdot e_1 \|^2_{L^2}, \| V \cdot e_2 \|^2_{L^2}, \ldots, \| V \cdot e_n \|^2_{L^2} \right) \in \R^{n}.
$$
More precisely, $T$ maps any orthonormal basis into the $(n-1)$-dimensional simplex
$$
\Delta^{n-1} = \{ x \in \R^n : x_1,\ldots,x_n \ge 0, \ x_1+\ldots+x_n = 1 \}. 
$$
Our claim is that $T$ admits the point $p_0 := (\frac 1n,\ldots,\frac 1n)$ as its value.
For $n=2$, this is easily seen by fixing an arbitrary orthonormal basis $e_1,e_2$ and rotating both vectors until we obtain $e_2,-e_1$. Since $\Delta^1$ is just a segment and points $T(e_1,e_2)$ and $T(e_2,-e_1)$ lie on the opposite sides of $p_0$, by continuity the value $p_0$ is also achieved.
For higher dimensions, let us assume that $T$ doesn't take the value $p_0$. Now the domain of $T$ is the set of all orthonormal bases, which can be identified with $O(n)$ (space of all orthogonal transformations). Thus, the domain is a compact set and of course $T$ is continuous, which means that the function
$$
(e_1,\ldots,e_n) \mapsto |T(e_1,\ldots,e_n)-p_0|^2 \in \R
$$
achieves a minimum. Let us consider such minimal basis $e_1,\ldots,e_n$. Of course, $T(e_1,\ldots,e_n) \neq p_0$, and so $\| V \cdot e_i \|^2_{L^2} \neq \| V \cdot e_j \|^2_{L^2}$ for some $i \neq j$. Applying the $2$-dimensional case, we can replace $e_i,e_j$ by an orthonormal basis $\hat{e}_i,\hat{e}_j$ of the same subspace $\operatorname{span}(e_i,e_j)$, satisfying
$$
\| V \cdot \hat{e}_i \|^2_{L^2} + \| V \cdot \hat{e}_j \|^2_{L^2} 
= \| V \cdot e_i \|^2_{L^2} + \| V \cdot e_j \|^2_{L^2}, 
\qquad
\| V \cdot \hat{e}_i \|^2_{L^2} = \| V \cdot \hat{e}_j \|^2_{L^2}.
$$
By strict convexity of the function $x \mapsto (x-\frac 1n)^2$, the pair $(\| V \cdot \hat{e}_i \|^2_{L^2},\| V \cdot \hat{e}_j \|^2_{L^2} )$ is strictly closer to $(\frac 1n,\frac 1n)$ than the pair $(\| V \cdot e_i \|^2_{L^2},\| V \cdot e_j \|^2_{L^2} )$. Considering the whole basis of $E$ with $e_i,e_j$ replaced by $\hat{e}_i,\hat{e}_j$, we obtain a contradiction with minimality of $e_1,\ldots,e_n$.
