diffeomorphism preserve a volume form

Question

Let $\omega_1$, $\omega_2$ two volume form on a compact manifold $M$, we know that there exists a never-vanishing function $f$, s.t. $\omega_1=f\omega_2$. If $h$ is a diffeomorphism $M \to M$ preserves $\omega_1$, does it preserve $\omega_2$?

Answer 1

No.

As a simple counterexample, consider the torus $T^2 = \mathbb{R}^2/\mathbb{Z}^2$ (I suppose it would be just as easy, or even easier, to think of a counterexample on $S^1$). Let $\omega_1 = dx \wedge dy$ be the volume form induced from the flat metric. Let $f$ be your favorite positive smooth function on $T^2$ (e.g., $f(x,y) = \sin^2(\pi x) + 1$), and let $\omega_2$ satisfy $\omega_1 = f \omega_2$. Let $h$ be a translation in the $x$-direction, say $x \mapsto x + \frac{1}{2}$. Then $h$ preserves $\omega_1$, but chances are, $h$ does not preserve $\omega_2$.

I suppose one could say that if $\omega_1 = f \omega_2$ and $h$ preserves both $\omega_1$ and $f$ (i.e., if $h^*f(x) := f(h(x))$ is equal to $f(x)$), then $\omega_2$ is also preserved by $h$.