Let $G \circlearrowright G$ be a Lie group acting on itself by conjugation: $\varphi_g(h) \doteq ghg^{-1}$. Denote by $\varphi^h$ the other partial map, that is, $\varphi^h(g) = ghg^{-1}$. I take as the definition of the infinitesimal generator of the action associated to $u \in \mathfrak{g}$ the vector field $(u^\#)_h = {\rm d}(\varphi^h)_e(u)$, where $u$ is the identity of $G$.
I have checked that for the action given by left translation, $u^\#$ is the unique right-invariant extension of $u$, and for the action given by right translation we have that $u^\#$ is minus the unique left-invariant extension of $u$.
I do not know how to compute $u^\#$ in this case, though. If $\alpha$ is a curve on $G$ with $\alpha(0) = e$ and $\alpha'(0) = u$, I get stumped at $$ (u^\#)_h = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \alpha(t)h\alpha(t)^{-1},$$since I can't just "factor" something like $L_h$ or $R_h$ like in the previous cases. Help?