Let $M$ be a smooth manifold, and $G$ a Lie group with Lie algebra $\mathfrak{g}$. The Lie algebra of the diffeomorphism group of $M$ is the Lie algebra of vector fields on $M$; that is $\text{Lie}(\text{Diff}(M)) = \Gamma(TM)$.

Given a smooth left-action of $G$ on $M$ expressed as a Lie group homomorphism $\lambda : G \to \text{Diff}(M)$, we obtain a Lie algebra homomorphism $\text{Lie}(\lambda): \mathfrak{g} \to \Gamma(TM)$.

What is the relation between this Lie algebra homomorphism and the fundamental vector field mapping $\zeta: \mathfrak{g} \to \Gamma(TM)$, which is an anti-homomorphism of Lie algebras? Recall that the fundamental vector field mapping is given by $$\zeta(X)_x = T_e \lambda(-, x).X$$ for any $x \in M$ and $X \in \mathfrak{g}$. Here I have written the group action as a smooth map $\bar{\lambda}: G \times M \to M$.

I don't know much about infinite-dimensional Lie groups and manifolds, personally, but it seems that there should be some simple relationship between these two mappings.