$\newcommand{\Cof}{\text{cof}}$ Let $d>2$. Let $f \in W^{1,p}(\Omega,\mathbb{R}^d)$ where $\Omega$ is an open subset of $\mathbb{R}^d$. Let $2 \le k \le d-1$ be fixed.
Suppose that $\det df>0$ a.e. and that $\bigwedge^k df$ is smooth. Is $f$ smooth?
Partial answer: If $k,d$ are not both even and $\bigwedge^k df \in \text{GL}(\bigwedge^{k}\mathbb{R}^d)$, the answer is positive. (see details below).
I am not sure the answer remains positive when $k,d$ are both even, since in that case, $\bigwedge^k A=\bigwedge^k (-A)$ and both $A,-A \in \text{GL}^+(\mathbb{R}^d)$. Thus the minors cannot distinguish beween a map and its negative, so theoretically $df$ could "switch" between "something" and its negative, thus violating smoothness.
It would be interesting to find a concrete counter example for smoothness in this case. The smallest dimensions are $d=4,k=2$. Every possible counter-example must have non-continuous weak derivatives.
When $k,d$ are not both even, $\bigwedge^k df$ uniquely determines $df$ (assuming $\det df>0$). That is, if $A,B \in \text{GL}^+(\mathbb{R}^d)$ and $\bigwedge^k A=\bigwedge^k B$, then $A=B$. Indeed, write $S=AB^{-1}$. Then we have $\bigwedge^k S=\text{Id}_{\bigwedge^k \mathbb{R}^d}$. This implies, any $k$-dimensional subspace of $\mathbb{R}^d$ is $S$-invariant, hence $S$ is a multiple of the identity, i.e. $S=\lambda \text{Id}$, which then forces $\lambda^k=1$, so $\lambda=\pm 1$. If $k$ is odd, then $\lambda=1$, and $S=\text{Id}$. If $d$ is odd, then the requirement $S \in \text{GL}^+(\mathbb{R}^n)$ forces $\lambda^d=\det S >0$, so again $S=\text{Id}$.
We showed that the map $\psi: A \to \bigwedge^k A$ is a smooth injective homomorphism of Lie groups $\text{GL}^+(V) \to \text{GL}(\bigwedge^{k}V)$. Set $ S=\text{Image} (\psi)$. Since $S$ is an embedded submanifold, $\psi:\text{GL}^+(V) \to S$ is a diffeomorphism. Now composing $x \to \bigwedge^k df_x$ with the smooth inverse of $\psi$ finishes the job.
Some more details are provided in my answer below.
