Lie derivative and preserving flows Let $\Omega$ be a volume form on a manifold $M$. Consider $X$ a vector field on $M$ which is not zero at any point. Is the following assertion true? 

If $ \mathcal{L}_{X}\Omega = 0$, then in a neighborhood of any point there exists a local chart $(x^{i})$ such that
  $$ X(x^{i}) = Const.$$
  $$ \Omega = dx^{1} \wedge ... \wedge dx^{n}.$$

Thanks in advance.
 A: Yes, there always exists such a chart. Here's a proof. Suppose $\mathcal L_X\Omega \equiv 0$ and $X$ is nowhere zero. Then the "straightening theorem" shows that for any $p\in M$, there are local coordinates $(y^i)$ centered at $p$ in which $X = \partial/\partial y^1$.  
Now we consider two cases based on $n=\text{dim}(M)$. First, if $n=1$, then we can write $\Omega = f(y^1)\,dy^1$ for some smooth function $f$, and $\mathcal L_X\Omega=0$ implies $f$ is constant in a neighborhood of $p$; let's say $f(y^1) \equiv c$. If we define a new coordinate by $x^1 =c y^1$, then $\Omega = dx^1$ and $X(x^1) =c$.
On the other hand, if $n>1$, write $\Omega = f(y^1,\dots,y^n)dy^1\wedge\dots\wedge dy^n$, and note that $\mathcal L_X\Omega=0$ implies that $f$ is independent of $y^1$ in a neighborhood of $p$: $f(y^1,\dots,y^n) = \hat f(y^2,\dots,y^n)$ for some smooth function $\hat f$ of $n-1$ variables.  Define new coordinates $(x^i)$ by
\begin{align*}
x^i(y^1,\dots,y^n) &= y^i, \qquad 1\le i \le n-1,\\
x^n(y^1,\dots,y^n) & = \int_0^{y^n} \hat f(y^2,\dots,y^{n-1},u)\,du.
\end{align*}
Then 
\begin{equation*}
\frac{\partial x^n}{\partial y^n} = \hat f(y^2,\dots,y^n),
\end{equation*}
and we conclude
\begin{equation*}
dx^j = dy^j \text{ for $1\le j\le n-1$}, \qquad 
dx^n = \hat f dy^n + \text{terms involving $dy^1,\dots,dy^{n-1}$}.
\end{equation*}
It follows easily that 
\begin{equation*}
dx^1\wedge\dots\wedge dx^n = \hat f\, dy^1\wedge \dots \wedge dy^n = \Omega.
\end{equation*}
On the other hand, 
\begin{equation*}
X = \frac{\partial}{\partial y^1 } = \sum_{j=1}^n \frac{\partial x^j}{\partial y^1}\, \frac{\partial}{\partial x^j} = \frac{\partial}{\partial x^1},
\end{equation*}
since $x^1 = y^1$ and  $x^2,\dots,x^n$ are independent of $y^1$. Therefore $X(x^1) \equiv 1$ and $X(x^j) \equiv 0$ for $j=2,\dots,n$.
