Group actions on manifolds - exponential map Let $M$ be a smooth manifold. Suppose $K$ is a Lie group (with Lie algebra $\mathfrak{k}$) acting EDIT: TRANSITIVELY on $M$ from the left and $G$ is a Lie group (with Lie algebra $\mathfrak{g}$) acting on $M$ from the right. Suppose further these actions commute - ie $k(pg)=(kp)g=kpg$. Fix a point $x\in M$. Suppose $\phi(t)$ is some smooth curve in $M$ such that $\phi(0)=x$.
Is it possible to find $X\in \mathfrak{k}$ and $A\in\mathfrak{g}$ such that
\begin{align*}
\frac{d}{dt}\bigg|_{t=0}\exp(tX)x\exp(tA)=\phi^{\prime}(0)??
\end{align*}
And if so, how could you prove such a fact?

What if the action of $G$ and $K$ are both trivial? E.g. $k\cdot x=x=x\cdot g$ for all $k$ and $g$? Then your expression $\mathrm{exp}(tX)\,x\, \mathrm{exp}(tA)$ is identically $x$, and so has trivial derivative. Hence you can't hit $\phi'(0)$ in general. 
