Let $f:M_1\longrightarrow M_2$ be a smooth map, Where $M_1, M_2$ are smooth Manifolds endowed with a Lie group $G$-action respectively (that is the same $G$ acts differently on the two manifolds), There is an infinitesimal version of each action, namely $$\mathfrak{g}\times M_i\longrightarrow TM_i$$ $$(X,m)\mapsto \underline{X}^i(m)$$ Where $\underline{X}^i$ is the fundamental vector field induced by $X\in\mathfrak{g}$ associated to the group action $G\times M_i\longrightarrow M_i$.
Now the map $f$ being equivariant under $G$ means for all $g\in G$ and $m\in M_1$ one has $$g^2\circ f (m)=f\circ g^1 (m)$$ Where I use $g^i$ to distinguish the different Actions.
And the $f$ being equivariant in the infinitesimal version means for all $X\in \mathfrak{g}$ one has $$\underline{X}^2(f(m))=(df)_{m}\underline{X}^1(m)$$
Then my question is, are these two statements equivalent? Well, it is obviously equivariant in the infinitesimal version if $f$ is equivariant, but how about the inverse? Can the infinitesimal thing decide the global one?
If $G$ is not connected then they are certainly not equivalent, since the infinitesimal statement only depends on the action of the connected component of the identity in $G$. For instance, if $G$ is $0$-dimensional, the infinitesimal statement is always trivially true.
If $G$ is connected, though, the infinitesimal statement does imply that $f$ is equivariant. To prove this, note that for $X\in\mathfrak{g}$, $t\mapsto \exp(tX)^i\cdot m$ is the flow along the vector field $\underline{X}^i$ starting from $m$. If $f$ is infinitesimally equivariant, then it preserves these flows, and so it preserves the action of elements of $G$ of the form $\exp(X)$. When $G$ is connected, it is generated by the image of the exponential map, and so it follows that $f$ is actually equivariant.