Let $S^3$ denote the unit quaternions. Let $L_g : S^3 \to S^3$ denote the left translation action for each $g \in S^3$.
Let $X : S^3 \to T_pS^3$ be the vector field given by $p\mapsto X_p =pi$, where $i= (0, 1, 0, 0)$. We say that $X$ is left invariant if $d(L_g)_{g'}(X_{g'})=X_{gg'}$ for any $g, g' \in S^3$, where $d(L_g)$ denotes the differential.
Since the quaternions $\mathbb H \cong \mathbb R^4$, it is not hard to verify that $X$ is left invariant when it is viewed as a vector field on $\mathbb H$ and $L_g$ acts on all of $\mathbb H$. But if we view it as a vector field on $S^3$, then technically we need to use local charts on the sphere to compute the differential and tangent vectors. This gets much messier.
My question is whether the fact that $X$ is left invariant when viewed on all of $\mathbb H$ implies it is left invariant when restricted to the submanifold $S^3$. If so, why?
This indeed implies that $X$ is also left invariant when restricted tot $S^3$. This is because the manifold $S^3$ is a Lie subgroup of $\mathbb{H}$ and because the immersion map (the identity) $id\colon S^3 \to \mathbb{H}$ is a group morphism.
Let $L_g\colon \mathbb{H}\to \mathbb{H}$ be the left translation on the whole of $\mathbb{H}$. Then the maps $L_g\circ id$ on $S^3$ and the map $L_g$ on $\mathbb{H}$ are the same. In particular the differential $d(L_g\circ id)_{g'}(X_{g'})=X_{gg'}$ on $S^3$ is the same as the differential $dL_g(X_{g'})=X_{gg'}$ on $\mathbb{H}$. (Basically you are just taking a restriction.)