For a lie group $G$ with bi-invariant metric, and left-invariant fields $X,U,V$, $\langle [U,X],V\rangle = -\langle U,[V,X]\rangle$

Specifically, I'm somewhat confused by the answer to this question here, which I reproduce below:

Let $g$ denote the metric of $G$. The right invariance of $g$ guarantees that $$R^*_{\exp (tX)}g=g, \forall t \in \mathbb{R}.$$ On the other hand, $t\to R_{exp(tX)}$ is the flow of X, we have $\mathcal{L}_Xg=0$, where $\mathcal{L}_Xg$ denotes the Lie derivative of $g$ in the direction X. $$(\mathcal{L}_Xg)(U,V)=X(g(U,V))-g(\mathcal{L}_XU,V)-g(U,\mathcal{L}_YZ) \\ =-\langle[X,U],V\rangle-\langle U,[X,V]\rangle$$

where we use the invariance of $g$ at last equal to conclude that $g(U,V)$, seen as function of $G$, is constant in each connected component of $G$.

My question is, why can we assume that $\mathcal{L}_X g = 0$?

Given a vector field $X$ with flow $\varphi_t^X$, the definition of the Lie derivative $\mathcal{L}_X g$ is
$$\mathcal{L}_X g = \frac{d}{dt} (\varphi_t^X)^{*}(g)|_{t = 0}.$$
In your case, $\varphi_t^X = R_{\exp(tX)}$ and the second line shows that $R_{\exp(tX)}^{*}g$ is constant (doesn't depend on $t$) so the derivative with respect to $t$ is zero.