The volume from on an oriented 4-dimensional (pseudo-) Riemannian manifold $(M,g)$ is given by $$\Omega :=\sqrt{\lvert g\rvert}\,dx^{0}dx^{1}dx^{2}dx^{3}:=\sqrt{\lvert g\rvert}\,d^{4}x$$ where $g=\text{det}(g_{\mu\nu})$ is the determinant of the metric tensor $g_{\mu\nu}$ (apologies for my sloppy physicist notation).
Given a diffeomorphism $\phi:M\rightarrow M$, the volume form $\Omega$ should change by a Lie derivative, i.e. $$\delta\Omega =\mathcal{L}_{X}\Omega$$ where $X$ is the vector field generating the diffeomorphism. Now I know that $$\delta\left(\sqrt{\lvert g\rvert}\right)=\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right) =\frac{1}{2}\sqrt{\lvert g\rvert}g^{\mu\nu}\delta g_{\mu\nu}=\sqrt{\lvert g\rvert}g^{\mu\nu}\nabla_{\mu}X_{\nu}$$ where I have used that $\delta g_{\mu\nu}=\mathcal{L}_{X}g_{\mu\nu}=\nabla_{\mu}X_{\nu}+\nabla_{\nu}X_{\mu}$. And so, so far I have $$\mathcal{L}_{X}\Omega=\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right)d^{4}x+\sqrt{\lvert g\rvert}\,\mathcal{L}_{X}\left(d^{4}x\right)=g^{\mu\nu}\nabla_{\mu}X_{\nu}\,\Omega+\sqrt{\lvert g\rvert}\,\mathcal{L}_{X}\left(d^{4}x\right)$$
However, I'm not sure how $d^{4}x$ transforms, so my question is: what is the Lie derivative of $d^{4}x$? i.e. what is $$\mathcal{L}_{X}\left(d^{4}x\right)\;?$$ I'm a physicist and so my knowledge of differential geometry is not that extensive, any help would be much appreciated.