Suppose that $M$ is a manifold with a fixed volume form $\mu$. I want to compute (at least heuristically, I know there are some delicate considerations for infinite manifolds that I am sweeping under the rug) the derivative of the map $f(\phi) = \phi_*\mu = (\phi^{-1})^*\mu$, where $\phi$ is a diffeomorphism of $M$. To this end, suppose that $(\phi_t)$ is a path of diffeomorphisms such that $\phi_0 = \phi$ and $\dot\phi_0 = v\circ\phi$ where $v$ is a vector field on $M$. I then start computing it, using the usual derivative of a pullback formula (e.g. Formula about time derivative of pushforward of family of forms: where is it from?): $$Df_{\phi}(v\circ\phi) = \left.\frac{d}{dt}\right|_{t=0}\psi_t^*(\mu) = \psi^*\left(\mathcal L_{X_0}\mu\right),$$ where I've abbreviated $\psi_t := \phi_t^{-1}$ and $(X_t)$ is a family of vector fields defined by: $$X_t = \left.\frac{d}{ds}\right|_{s=0}(\psi_{s+t}\circ\psi_t^{-1}).$$ My goal is now to compute $X_0$, which in my head I thought should be something like $-v\circ\phi$ (because of Lie group aspect of the diffeomorphism group), however I am having difficulties computing this. I thought it would be something like: $$X_0 = \dot\psi_0\circ\psi_0^{-1} = \psi_*\dot\psi_0,$$ and then we can compute $\dot\psi_t = -D(\phi_t^{-1})(\dot\phi_t)$ using something like Time derivative of inverse of flow diffeomorphisms, but then I get $\dot\psi_0 = -D\psi(v\circ\phi)$ and our expression becomes $X_0 = -\psi_*(D\psi(v\circ\phi))$.
Not thinking hard about it, I got confused between the "pushforward" that is $D\psi$ and the pushforward that is the pullback by the inverse, and made the naive computation: $$X_0 = -\psi_*(D\psi(v\circ\phi)) \approx -\psi_*\psi_*\phi^*v = -\psi_*v = -v\circ\phi\circ\phi\circ\phi,$$ But this is desperate thinking and I don't think it is remotely sane mathematics.
So my question is, can $X_0$ be simplified nicely to something in terms of $\phi$ and $v$?
Now I have looked at this a different way as follows and arrived at a similar conclusion that I wanted. Let $\phi_t$ be the path as described above (but all subsequent material ignored) and consider the isotopy $\psi_t := \phi_t\circ\phi^{-1}$. There exists a time dependent vector field $X_t$ such that $\frac{d}{dt}\psi_t = X_t\circ\psi_t$ (I force the isotopy because I have only seen this theorem apply to isotopies, but I am pretty sure what I do next suggests it holds for paths), and expanding out the definition of $\psi_t$ gives that $X_t = \dot\phi_t\circ\phi^{-1}_t$. Then the trick is that: $$ 0 = \frac{d}{dt}\mu = \frac{d}{dt}(\phi_t^*(\phi_t)_*\mu) = \phi_t^*\left(\frac{d}{dt}(\phi_t)_*\mu + \mathcal L_{X_t}((\phi_t)_*\mu)\right)$$ and taking the pushforward on both sides gets you that: $$Df_{\phi}(v\circ\phi) = \left.\frac{d}{dt}\right|_{t=0}(\phi_t)_*(\mu) = -\mathcal L_{X_0}(\phi_*\mu),$$ and a quick computation yields that $X_0 = \dot\phi_0\circ\phi_0^{-1} = v$, giving $Df_{\phi}(v\circ\phi) = -\mathcal L_{v}(\phi_*\mu)$.
Is there any way to get the same result using the reasoning from above?
