Show that $\Phi_t^*\omega=\omega \quad \forall t\iff (\mathrm{div}_{\omega}\mathbb{X})\omega=0$

Let $\omega$ be a non-vanishing differential $n$-form on $\mathbb{R}^n$, let $\mathbb{X}$ be a smooth vector field on $\mathbb{R}^n$ and define the divergence of $\mathbb{X}$ with respect to $\omega$ as $$L_{\mathbb{X}}\omega=(\mathrm{div}_{\omega}\mathbb{X})\omega.$$

I want to show that $$\Phi_t^*\omega=\omega \quad \forall t\iff (\mathrm{div}_{\omega}\mathbb{X})\omega=0$$ where $\Phi_t$ is the flow of $\mathbb{X}$ and $*$ denotes the pullback.

I think this suffices for one direction:

Suppose $\Phi_t^*\omega=\omega$. Then

\begin{align} (\mathrm{div}_{\omega}\mathbb{X})\omega & = L_{\mathbb{X}}\omega \\ & =\left. \frac{\partial}{\partial t} \Phi_t^* \omega \right|_{t=0} \\ & = \left. \frac{\partial}{\partial t}\omega \right|_{t=0} \\ & = 0 \end{align}

Now for the other direction, suppose we have $(\mathrm{div}_{\omega}\mathbb{X})\omega=0$. So $$\left. \frac{\partial}{\partial t} \Phi_t^* \omega \right|_{t=0}=0$$

I'm not really sure what to do with this part.

$\left. \frac{\partial}{\partial t} \Phi_t^* \omega \right|_{t=0}=0$ implies that the function $f(t):t\rightarrow \phi_t^*\omega$ is constant and in particular $f(t)=f(0)=\phi_0^*\omega=\omega$ since $\phi_0=Id$.
To say that $$f:t\mapsto \phi_t^*\omega$$ is constant you actually need to show that $$\partial_t\phi_t^*\omega=0$$ for all $$t$$, not just for $$t=0$$.
To do this, you really need to use properties of the flow of a vector field, in particular $$\phi_t\circ\phi_s=\phi_{t+s}$$. Indeed, using this fact we find: $$\partial_t\phi^*_t\omega=\frac{\mathrm{d}}{\mathrm{d}s}\Bigr|_{s=0}\phi^*_{t+s}\omega=\frac{\mathrm{d}}{\mathrm{d}s}\Bigr|_{s=0}\phi^*_{s+t}\omega=\phi^*_t\left(\frac{\mathrm{d}}{\mathrm{d}s}\Bigr|_{s=0}\phi^*_{s}\omega\right)=\phi^*_t\left(\mathcal{L}_X\omega\right)=0.$$
From here we can continue as in Tsemo Aristide's answer, i.e. since the function $$f$$ is constant and $$f(0)=\omega$$ we deduce that $$f(t)\equiv\omega$$.