In chapter 2 of this book (entitled: The Simplicity Of Diffeomorphism Groups) the author says that given any compactly supported smooth vector field $V$ on a simply connected and connected (finite-dimensional) manifold $M$, the corresponding flow map is stable under conjugation by compactly-supported diffeomorphisms.
Namely, let $F^V$ be map sending $x$ to $f(1)$ where $f$ solves the ODE (locally on some chart): $$ \frac{\partial_t f}{\partial t} = V(f(t)) \, f(0)=x. $$ Let $\phi:M\to M$ be a compactly supported diffeomorphism of $M$ (ie: outside some compact $\phi$ is the identity map) then $$ \phi^{-1}\circ f\circ \phi, $$ is again a solution to an ODE above? Why is this the case? $f\circ \phi$ is clear but I don't see why $\phi^{-1}$ again gives a flow....
Precomposition by $\phi$ just changes the initial point of your flow so I'll leave that to you. For the rest, we need that $\phi^{-1}$ is at-least $C^1$, which we have in your case. Let's argue locally so, WLOG, we identify an open nhd of $f\circ \phi(x)$ with $\mathbb{R}^d$.
By the chain-rule we have that $$ (\phi^{-1}\circ f)'(t) = (\phi^{-1})'(f(t))f'(t) = (\phi^{-1})'(f(t))V(f(t)) = \tilde{V}(f(t)). $$ Where $\tilde{V}(x)=(\phi^{-1})'(x)V(x)$ and the product is continuous since $\phi^{-1}\in C^1$ and $V$ was $C^1$ by your hypothesis.
So $ \phi^{-1}\circ f\circ \phi $ is again a flow!