Let $f,g:\mathbb{R}^m \to \mathbb{R}^n$ be two smooth functions and let $k$ be a strictly positive integer. Write $f \sim\_k g$ if at each point in the domain, the determinants of all $k \times k$ minors of the Jacobians of $f$ and $g$ coincide. This is clearly an equivalence relation; and more generally, one could let $f$ and $g$ be smooth maps of smooth manifolds $X \to Y$, and ask that for each $k$-form $\omega$ on $Y$ the pullbacks $f^\*\omega$ and $g^\*\omega$ agree as $k$-forms on $X$.


Have such equivalences been studied and given an alternate characterization? Here's the sort of thing I'm looking for: if $k=1$ and we are mapping between Euclidean spaces, $f$ and $g$ must lie in the same orbit of the translation group acting on the codomain. Perhaps there is a similar Lie-theoretic reformulation of $\sim\_k$ for maps $X \to Y$ even when $k > 1$?



Equivalence generated by Jacobian minors
First, let me point out that the OP's suggested generalization to arbitrary target manifolds $Y$ of asking that $f^\*\omega = g^\*\omega$ for *all* $k$-forms on $Y$, is *not* equivalent to the question about equality of $k$-by-$k$ minors.


To see this, consider the simplest case, $X = Y=\mathbb{R}^1$ and $k=1$. The general $1$-form is of the form $\omega = h(x)\,\mathrm{d}x$ (where $x:\mathbb{R}\to\mathbb{R}$ is the identity function), and $f^\*\omega = g^\*\omega$ for all $1$-forms $\omega$ is equivalent to
$$
h\bigl(f(x)\bigr)f'(x) = h\bigl(g(x)\bigr)g'(x)
$$
for all functions $h:\mathbb{R}\to\mathbb{R}$, which clearly implies that $f(x) = g(x)$ for all $x\in\mathbb{R}$. Meanwhile, $f^\*\mathrm{d}x = g^\*\mathrm{d}x$ only implies $f'(x) = g'(x)$, which *is* the same as requiring that all the $1$-by-$1$ Jacobian minors of $f$ and $g$ are equal, which does imply that they differ by an additive constant.


Instead, I think that the natural generalization is that one endows the $n$-manifold $Y$ with a *coframing*, i.e., a basis $\omega = (\omega^1,\ldots,\omega^n)$ of the $1$-forms on $Y$ (which, of course, requires that $Y$ be parallelizable, in fact, $\omega:TY\to\mathbb{R}^n$ defines a linear isomorphism $\omega\_y:T\_yY\to\mathbb{R}^n$ for each $y\in Y$) and then require that
$$
f^\*(\omega^{i\_1}\wedge\cdots\wedge\omega^{i\_k}) 
= g^\*(\omega^{i\_1}\wedge\cdots\wedge\omega^{i\_k})
$$
for all $i\_1<i\_2<\cdots <i\_k$.


Even this will generally force some 'finite' equations on $f$ and $g$ if $\omega$ is chosen generally (and the dimension of $X$ is greater than $k$). For a multi-index $I = (i\_1,\ldots,i\_p)$ with $1\le i\_1<i\_2<\cdots<i\_p\le n$, write $|I|=p$ and $\omega^I$ for $\omega^{i\_1}\wedge\cdots\wedge\omega^{i\_p}$. Then there will be functions $h^I\_J$ on $Y$ such that
$$
\mathrm{d}\omega^I = \sum\_{|J|=|I|{+}1} h^I\_J\,\omega^J.
$$
If the functions $h^I\_J$ are not constant, and $f$ and $g$ both have differential rank at least $k{+}1$, then applying the exterior derivative to the equations $f^\*\omega^I = g^\*\omega^I$ for $|I|=k$ will generally force some relations on the functions $f^\*h^I\_J$ and $g^\*h^I\_J$. One can avoid this 'problem' by assuming that the functions $h^I\_J$ be constants. For example, when one takes the standard coordinate coframing on $Y = \mathbb{R}^n$, one has $h^I\_J=0$. More generally, if $Y$ is a Lie group and the $\omega^i$ are a basis for the left-invariant forms on $Y$, then the $h^I\_J$ are constants. In this latter case, when $X$ is connected, one will have $f^\*\omega^i=g^\*\omega^i$ for all $i$ if and only if $g = \lambda\_y\circ f$ where $\lambda\_y:Y\to Y$ is left multiplication by $y\in Y$ (regarded as a Lie group). So I think that this is the natural generalization of the OP's case of $\mathbb{R}^n$.


Second, let me point out that, if the differential ranks of $f$ and $g$ are both less than $k$ at every point, then, of course, $f^\*\omega = g^\*\omega=0$ for all all $k$-forms $\omega$ on $Y$, so there is no further condition implied than the rank condition. Thus, to get an interesting problem, one must assume that the differential ranks of $f$ and $g$ are *at least* $k$ in order to get an interesting theory.


Once one assumes this, there are some reasonable things to say. For example, if one assumes that the differential ranks of $f$ and $g$ are both at least $k$, then the condition $f^\*\omega^I = g^\*\omega^I$ for all $|I|=k$ implies that $\mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T\_xX$ and $\omega\_x\bigl(f'(x)(T\_x)\bigr) = \omega\_x\bigl(g'(x)(T\_x)\bigr) \subset\mathbb{R}^n$ for all $x\in X$. Moreover, if one sets $K\_x = \mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T\_xX$ and $Q\_x = \omega\_x\bigl(f'(x)(T\_x)\bigr) = \omega\_x\bigl(g'(x)(T\_x)\bigr) \subset\mathbb{R}^n$, then the induced isomorphisms
$$
[f'(x)],[g'(x)]:T\_xX/K\_x\to Q\_x
$$
satisfy, for all $x\in X$,
$$
E\_k\bigl([f'(x)]\bigr) = E\_k\bigl([g'(x)]\bigr):E\_k(T\_xX/K\_x)\to E\_k(Q\_x), 
$$
where $E\_k$ is the '$k$th-exterior power' functor on the category of vector spaces and linear maps.


When the differential ranks of $f$ and $g$ are equal to $k$, this is not a very strong condition, so $f$ and $g$ need not be closely related. For example when $m=n=k$, and $X = Y = \mathbb{R}^k$, then the OP's condition on $f$ and $g$ reduces to the assumption that the $f$- and $g$- pullbacks of the standard volume form on $Y$ are equal, and, of course, there are many such pairs $f$ and $g$ besides the translations.


When the differential ranks of $f$ and $g$ are both greater than $k$, though, this is a much stronger condition. In fact, one finds that, when $k$ is odd, this is equivalent to $[f'(x)] = [g'(x)]$ while, when $k$ is even, this is equivalent to $[f'(x)] = \pm[g'(x)]$. Once one is in this situation, at least when one assumes that the differential ranks of $f$ and $g$ are constant, the Cartan equivalence method can be applied. For example, one has the following result:


**Proposition:** Suppose that $f,g:\mathbb{R}^m\to\mathbb{R}^n$ are smooth maps of constant differential rank *greater* than $k\ge1$ and suppose that all their corresponding $k$-by-$k$ Jacobian minors are equal. If $k$ is odd, then $g = c + f$ where $c\in\mathbb{R}^n$ is a constant. If $k$ is even, then $g = c \pm f $ where $c\in\mathbb{R}^n$ is a constant.


