= Solution
Use the following permitted kernel result, stated in the form needed here. For an <Anosov flow> generated by a smooth <surface thermostat> on a closed oriented <Riemannian surface>, if a smooth function $u$ on $SM$ satisfies
$$
Fu=h\circ\pi+\eta_x(v),
$$
where $h$ is a smooth base function and $\eta$ is a smooth one-form, then $h=0$ and $\eta$ is an <exact differential form>. Conversely these two conditions give a smooth solution by taking a base primitive. This is the <kernel of the thermostat X-ray transform in degrees zero and one>. In particular it applies to <Gaussian thermostats> without an additional curvature condition; it is stronger than the magnetic or geodesic special case.
Put $\theta=E^\flat$, so $\theta_x(v)=\langle E(x),v\rangle$. Suppose the thermostat preserves a smooth <volume form>. Write it as $\rho\mu$ with smooth $\rho>0$, choosing its sign separately on each component if necessary. By the <invariant volume criterion for a smooth flow> and the divergence computation in part (b),
$$
0=\operatorname{div}_{\rho\mu}F=-\theta_x(v)+F\log\rho.
$$
Thus $u=\log\rho$ is a smooth solution of $Fu=\theta_x(v)$. Apply the stated kernel theorem with $h=0$ to obtain $\theta=df$ for a smooth base function $f$. The definition of the <gradient> then gives $E=\operatorname{grad}_g f$.
Conversely, suppose $E=\operatorname{grad}_g f$. Since $d\pi(F)=v$, the vertical part has no effect on a base function and
$$
F(f\circ\pi)=df_x(v)=\langle E(x),v\rangle.
$$
The <divergence of a Gaussian thermostat> and the volume-change formula therefore give
$$
\operatorname{div}_{e^{f\circ\pi}\mu}F
=-\langle E,v\rangle+F(f\circ\pi)=0.
$$
Consequently
$$
\boxed{E=\operatorname{grad}_g f
\iff\phi_t\text{ preserves a smooth volume form},
\qquad \Omega=e^{f\circ\pi}\mu\text{ is invariant}.}
$$
Only the necessary direction uses the Anosov kernel theorem; the displayed invariant density works for any potential <Gaussian thermostat>. In particular its exponent is positive, as required to cancel the negative divergence found in part (b).
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