Suppose $X\in \mathfrak{g}$ is a left invariant vector field on a Lie group G. In this article it mentions that
The fact that our vector fields satisfy $L^*_gX = X$ implies that the flow commutes with left-translation: $\Phi_t\circ L_g = L_g \circ \Phi_t$.
It makes intuitive sense to me why this would be true, but I can't seem to formulate it. Let $h\in G$, then we want to show $\Phi_X^t(L_g(h)) = g\cdot \Phi^t_X(h)$. We have $X_{h} = X_{L_{g^{-1}}(gh)}= L_{g^{-1}}^*(X_{gh})$, but now I am getting lost again...
$\newcommand{\dt}{\frac{d}{dt}}$The strategy here is to show that $L_g \Phi_t L_g^{-1} = \Phi_t$ using the uniqueness of flow. Thus we differentiate the LHS at a point $p$ and time $t$, and replace the $\Phi_t$ with the flow of $L_g^* X$ (since the assumption is $L_g^* X = X$):
$$ \dt (L_g \Phi_t L_g^{-1} (p)) = dL_g (\dt \Phi_t (L_g^{-1} p)) = dL_g (L_g^* X(\Phi_t L_g^{-1} p)).$$
Now, since $L_g^*X (h) = dL_g^{-1}(x(L_g h))$, this is simply $X(L_g\Phi_t L_g^{-1}p)$; i.e. we have shown $$\dt (L_g \Phi_t L_g^{-1} (p)) = X(L_g\Phi_t L_g^{-1}p).$$ Since $L_g\Phi_0L_g^{-1}$ is the identity, this says exactly that $L_g \Phi_t L_g^{-1}$ is the flow of $X$.