I was wondering whether left invariance of a vector field $X$ to a respective Lie group $G$ (so $dL(a)(x)(X(x))= X(ax))$ is transfered to the respective flow defined by $\frac{d}{dt} \phi^{t}(x) = X(\phi^{t} (x))$ in the sense that
$$a \phi^{t}(x) = \phi^{t}(ax)?$$
This is something that sounds natural, but I don't know if it is true. Does anybody know a proof or a counterexample?
If anything is unclear, please let me know.
I'm not an expert in this subject, so pardon if I've made any mistakes, but I believe the answer is yes.
$\phi^t(ax)$ is the integral curve which passes through $ax$ at $t = 0$ with velocity $X(ax)$. To check that $a \phi^t(x)$ is equal to this integral curve, it suffices to check that it is an integral curve which passes through $ax$ at $t = 0$ (which it does), by the uniqueness of solutions. So we only have to calculate:
$\displaystyle\frac{d}{dt} a \phi^t(x) \Bigr|_{t = t_0} = \frac{d}{dt} L_a(\phi^t(x)) \Bigr|_{t = t_0} = d_{\phi^{t_0}(x)} L_a(X(\phi^{t_0}(x))) = X(a \phi^{t_0}(x))$