Why the left invariant means $L_y\circ x_t = x_t \circ L_y$?

Question

Consider a Lie group $G$, $$ L_x : G\rightarrow G, ~~~~~ L_x(y)=xy $$ is the translations from the left.

A vector field $X$ on $G$ is left invariant if $$ dL_x (X)=X. $$ Let $x_t$ be the flow of $X$, then why we have $L_y\circ x_t = x_t \circ L_y$ ?

There is a similar question. But I can't understand it. The answer hint

Let $\theta^{(p)}(t) = yx_t(p)$ and $\psi^{(p)}(t) = x_t(yp)$. Compute both ${\theta^{(p)}}'(0)$ and ${\psi^{(p)}}'(0)$. Then use uniqueness of integral curves.

But in my calculation,
$$ {\theta^{(p)}}'(0) =\frac{d}{dt}|_{t=0} ~yx_t(p)=yX(p) \\ {\psi^{(p)}}'(0)= \frac{d}{dt}|_{t=0} x_t(yp) = X(yp) $$ Evenly, I don't know what is the $yX(p)$. Since $y$ is a element of $G$, and $X(p)$ is a tangent vector at $p$, how do they multiply ?

Answer 1

I dealt with something similar a few days ago, so I figure it'd be helpful to write out this argument out in full. Given $X\in\mathfrak{X}(G)$ left-invariant, with flow $x_t$, you want to prove that $L_y\circ x_t=x_t\circ L_y$ for any $y\in G$. Let's fix a point $p\in G$. Both sides give $yp$ at $t = 0$. Differentiating the left at $t=0$ gives you $$ \left.\frac{d}{dt} \right|_{t=0} L_y(x_t(p)) = d(L_y)_p(X_p) = X_{yp} $$ by left-invariance of $X$, and differentiating the right at $t = 0$ gives you $X_{yp}$. So $t \mapsto (L_y \circ x_t)(p)$ and $t\mapsto (x_t \circ L_y)(p)$ are both integral curves of $X$ through $yp$. By uniqueness of integral curves, they agree for all $t$, and since $p\in G$ was arbitrary, $L_y \circ x_t = x_t \circ L_y$.

Or, you can say that this is a consequence of the more general result that if $X \in \mathfrak{X}(M)$ and $Y\in\mathfrak{X}(N)$ are $f$-related, for a smooth map $f\colon M \to N$, then $f$ takes integral curves of $X$ to integral curves of $Y$. The proof is almost the exact same. See, for example, Proposition 9.6 in Lee's Introduction to Smooth Manifolds.