Consider two vector-valued functions $f$ and $g$ of a vector $x$. If the jacobians of these two functions with respect to $x$ are exactly transposes of each other, what can we say directly about the relationship between $f$ and $g$? Specifically:
$$ x \in \mathbb{R}^n,\ \ \ f: \mathbb{R}^n \to \mathbb{R}^n,\ \ \ g: \mathbb{R}^n \to \mathbb{R}^n$$
$$ \nabla f := \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \dots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \dots & \frac{\partial f_n}{\partial x_n}\end{bmatrix},\ \ \ \nabla g := \begin{bmatrix} \frac{\partial g_1}{\partial x_1} & \dots & \frac{\partial g_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial g_n}{\partial x_1} & \dots & \frac{\partial g_n}{\partial x_n}\end{bmatrix}$$
$$\text{Suppose:}\ \ \nabla f = (\nabla g)^T\ \ \forall x\ \ $$
Using this relationship between the derivatives of $f$ and $g$, can we express any relation between $f$ and $g$ themselves?
Worded another way: given $f(x)$ and its jacobian $\nabla f$, can you find a function $g(x)$ with jacobian equal to $(\nabla f)^T$? Presumably you could find it in terms of $f(x)$? At first I thought it would just be some simple rearrangement of variables, perhaps $f$ of a permutation of $x$, but the actual relation is proving to be rather complicated.
Example:
\begin{align} f(x) = \begin{bmatrix} x_1 x_2 \\ x_1^2 + x_2^3 \end{bmatrix}\ \ \ &\implies\ \ \ \nabla f = \begin{bmatrix} x_2 & x_1 \\ 2x_1 & 3x_2 \end{bmatrix} \\ g(x) = \begin{bmatrix} g_1(x) \\ g_2(x) \end{bmatrix}\ \ \ &\implies\ \ \ \nabla g = \begin{bmatrix} x_2 & 2x_1 \\ x_1 & 3x_2 \end{bmatrix} \end{align}
Is there a general form for all of these special function-pairs $(f, g)$ which have transposed jacobians? Even simple examples are, in general, complicated PDE's to actually solve, but I'm hoping there is some clever way to think about what having the transpose of a given jacobian would mean, perhaps geometrically.