dg.differential geometry - Does a smooth, constant-rank, integrable distribution have a basis in which the traces of the structure constants are the divergences of the corresponding basis elements?

In a previous question, I asked an utterly trivial question, which Deane Yang correctly pointed out was utterly trivial. I will now ask a similar question, which is the one I meant to ask last time; I hope it's not similarly trivial.

I am working on $mathbb R^n$, although in fact any manifold with volume form is good enough. And my question is local: I have an open neighborhood $U ni 0$, and you are allowed to make it smaller if you want.

Recall that a constant-rank distribution $D$ on $U$ is a vector subbundle of the tangent bundle ${rm T}U$. Let's fix the rank to be $kleq n$, and suppose that everything is smooth: $Gamma(D)$ is a $C^infty$-submodule of $Gamma({rm T}U)$. The distribution is involutive if $Gamma(D)$ is a Lie subalgebra of $Gamma({rm T}U)$ (over $mathbb R$ — the Lie bracket on $Gamma({rm T}U)$ is not $C^infty$-linear). The distribution $D$ is smooth if $Gamma(D)$ is a free rank-$k$ module over $C^infty$, i.e. if $Gamma(D)$ has a basis ${v_1,dots,v_k}$ so that $Gamma(D) = operatorname{Span}_{C^infty}{v_1,dots,v_k}$.

So, suppose that on $U subseteq mathbb R^n$ I have a smooth involutive rank-$k$ distribution. Given a basis $v_1,dots,v_k in Gamma(D)$ (and I will use coordinates $x^1,dots,x^n$ on $mathbb R^n$, so I will write $v_a = sum_i v_a^i(x) frac{partial}{partial x^i}$), then I can define the structure coefficients $f_{a,b}^c(x)$, $a,b,c = 1,dots,k$, via $[v_a,v_b] = sum_c f_{a,b}^c v_c$, or, in coordinates:
$$ sum_i left( v_a^i(x),frac{partial v_b^j}{partial x^i} - v_b^i(x),frac{partial v_a^j}{partial x^i}right) = sum_c f_{a,b}^c(x),v_c^j(x) $$

Question: Does there exist a basis ${v_a}$ for a given smooth involutive constant-rank distribution so that for each $a=1,dots,k$ and each $xin U$, we have $displaystyle sum_b f^b_{a,b}(x) = sum_i frac{partial v^i_a(x)}{partial x^i}$?

For example, by Frobenius theorem (my utterly trivial question), I can find a basis so that the LHS vanishes for each $a$. Or, by another of my trivial questions, I could make the basis entirely divergence-free. But I don't think I can simultaneously make the basis consist of divergence-free vector fields.

Question: If so, how many choices of such a basis do I have? Clearly ${rm GL}(k,mathbb R)$ acts on the set of choices; are there more?