Is there a matrix that converts the gradient of any function to gradient of other function? The study of hamiltonian mechanics brought me to the following question.

Let $n$ be a natural number ($n&gt;1$).

Let $A(\mathbf{x})$ be a $n\times n$ matrix consisting of functions $a_{ij}(\mathbf{x})$ ($a_{ij}:\mathbb{R}^n\to\mathbb{R}$):
$$
A(\mathbf{x})=
\begin{pmatrix}
a_{11}(\mathbf{x})&amp; \cdots&amp; a_{1n}(\mathbf{x})\\
\vdots&amp;\ddots&amp;\vdots\\
a_{n1}(\mathbf{x})&amp;\cdots&amp; a_{nn}(\mathbf{x})
\end{pmatrix}.
$$
Let $A(\mathbf{x})$ be so, that for any $F(\mathbf{x})$ ($F:\mathbb{R}^n\to\mathbb{R}$):
$$
\begin{pmatrix}
a_{11}(\mathbf{x})&amp; \cdots&amp; a_{1n}(\mathbf{x})\\
\vdots&amp;\ddots&amp;\vdots\\
a_{n1}(\mathbf{x})&amp;\cdots&amp; a_{nn}(\mathbf{x})
\end{pmatrix}
\begin{pmatrix}
\frac{\partial F}{\partial x_1}\\
\vdots\\
\frac{\partial F}{\partial x_n}
\end{pmatrix}
=
\begin{pmatrix}
g_1(\mathbf{x})\\
\vdots\\
g_n(\mathbf{x})
\end{pmatrix}
=
\begin{pmatrix}
\frac{\partial G}{\partial x_1}\\
\vdots\\
\frac{\partial G}{\partial x_n}
\end{pmatrix}
$$
for some $G(\mathbf{x})$ ($G:\mathbb{R}^n\to\mathbb{R})$.

In other words, if we multiply fixed $A(\mathbf{x})$ by the gradient of any $F(\mathbf{x})$ we necessarily get the gradient of some $G(\mathbf{x})$.

Can we say anything about such $A(\mathbf{x})$? I would be glad if the only opportunity is that $A(\mathbf{x})=cE$, where $E$ is the identity matrix and $c$ is some real number. Is it correct? Is it possible to prove it?

All the functions are considered to be "good enough" ("smooth enough").
I have found the solution (with the help of Mathoverflow users).

See here: https://mathoverflow.net/questions/239275/is-there-a-matrix-that-converts-the-gradient-of-every-possible-function-to-gradi
I have found the solution (with the help of Mathoverflow users).

See here: https://mathoverflow.net/questions/239275/is-there-a-matrix-that-converts-the-gradient-of-every-possible-function-to-gradi