Let $v$ be a vector field. Does there exists a volume form $\Omega$ 
such that its Lie derivative is proportional to itself with a constant coefficient:
$$\mathcal{L}\_v \Omega= C \cdot \Omega? \ \ \ \ \ (\ast)$$


A simplification of the question: assume that the divergence $\sum\_i \frac{\partial v^i}{\partial x\_i}$
 of the vector field is not zero at all points (in this case $C$ is necessary not zero)? 


A related question: assume in addition that the divergence is zero, and require that the constant $C$ in $(\ast)$ is zero? 


The questions are local (i.e., we work in an arbitrarily small neighborhood), 
 everything is $C^\infty$-smooth, and is even real-analytic if it makes the life easier. The dimension is arbitrary. 


Of course near the points where the vector field does not vanish the existence of such a volume form follows from the existence of a coordinate system such that our vector field is $\partial\_{x\_1}$. More generally, if a vector field is linearisable near a point, then the existence of such a volume form is also trivial. 


Actually, I believe that the answer on the very first question in negative;
 this belief is because of the divergence of the vector field controls the coefficient $C$ and one can possibly build a counterexample by constructing a vector field such that it vanishes at a convergent sequence of points $a\_1,..., a\_k, ... \to a$, such that the divegence is zero at the point $a$ and is not zero at all 
the points $a\_k$. 


The motivation came from projective differential geometry: it is known (see for example 
[Projectively equivalent connections](https://mathoverflow.net/questions/122311/projectively-equivalent-connections/122334#122334)) that projective structure + a volume form up to a constant coefficient uniquely defines the affine structure. Thus, a positive answer on the very first question would imply that a projective vector field always preserves a affine connection, which would make the investigation of say the singular points of projective vector fields much easier.


Added after the answer and comment of Ben McKay: The comment, and then the answer of Ben McKay does answer two of three questions I pose. The remaining question that I do have hote to get an answer (and, hopefully, a positive one) is whether any vector field with nonzero divergence is homothety vector field for a volume form? 



Does for every vector field there always exist a volume form for which the vector field is a homothety?
If there is such a volume form $\Omega$, by Moser's theorem we can pick local coordinates in which $\Omega=dx^1 \wedge \dots \wedge dx^n$. If in some coordinates $v$ vanishes to order $k$, for some $k>1$, but not at order $k$, then the same is true in any coordinates. For example, we can suppose that $v=f^2w$ with $f=0$ at some point where $df \ne 0$ and $w \ne 0$ and $w$ not tangent to $f=0$ at that point. This condition is coordinate invariant, and we calculate that $\mathcal{L}\_v \Omega=C \Omega$ just when $C=0$ and $2\mathcal{L}\_w f + f \, \text{div} w = 0$. But at $f=0$ this forces $\mathcal{L}\_w f =0$ tangent, a contradiction. So $v$ does not preserve $\Omega$. But then $v$ does not preserve any volume form.


