The action of the group of invertible matrices on vector bundle and equivariance of certain isomorphism Let $G:=GL^+(n)$ be a group of invertible matrices with positive determinant, $K:=SO(n)$ and let $E=\mathfrak{g}/ \mathfrak{k}$ be a quotient of corresponding Lie algebras. We can form the vector bundle $(G \times E)/K$ where the action of $K$ on $G \times E$ is as follows:
$$a \cdot (g,v):=(ga,ad(a^{-1})v)$$
This bundle turns out to be isomorphic with the tangent bundle $T(G/K)$. This isomorphism is defined as $\pi_K(g,v) \mapsto (\pi(g),\pi_*L_{g*}v)$. 

I would like to understand how the group $G$ acts on $(G \times E)/K$ and knowing this: whether our isomorphism is equivariant.   

Related discussion is here Tangent bundle of the quotient of matrix Lie groups
EDIT: Sorry if my notation was confusing: here $L_g$ is a left multiplication map, lower star is a differential (in suitable point) and $\pi_K:G \times E \to (G \times E)/K$ and $\pi:G \to G/K$ are projections. $v$ was supposed to live already in $E$. 
 A: The action of $G$ on $(G\times E)/K$ is simply induced by left multiplication on the first factor. So writing the orbit of $(g,v)$ as $[(g,v)]$ you simply define $\tilde g\cdot [(g,v)]:=[(\tilde gg,v)]$. This is well defined since left and right translations in a group commute, so you get a left $G$-action on $(G\times E)/K$. 
To see that the isomorphism is equivariant, I'll slightly change your notation, since I don't completely understand it. Indeed the first component in your formula is redundant, since it is just the foot-point of the tangent vector in the sedcond component. I denote by $L_v$ the left invariant vector field generated by by $v\in\mathfrak g$, so $L_v(g)=T_e\lambda_g\cdot v$, where $\lambda_{g}:G\to G$ is left translation by $g$. So we have $\pi_K([(g,v+\mathfrak k)])=T_g\pi\cdot L_v(g)\in T_{gH}G/K$. Now we want to compute $\pi_K([\tilde gg,v+\mathfrak k)]=T_{\tilde gg}\pi\cdot L_v(\tilde gg)$. By left invariance, $L_v(\tilde gg)=T_g\lambda_{\tilde g}\cdot L_v(g)$ and by definition $\pi\circ\lambda_{\tilde g}=\ell_{\tilde g}\circ\pi$, where $\ell_{\tilde g}:G/K\to G/K$ denotes the left action of $\tilde g$. Differentiating this equation in the point $g$, we get $T_{\tilde gg}\pi\circ T_g\lambda_{\tilde g}=T_{\pi(g)}\ell_{\tilde g}\circ T_g\pi$. Inserting this above, we get 
$$
\pi_K([\tilde gg,v+\mathfrak k)]=(T_{\pi(g)}\ell_{\tilde g}\circ T_g\pi)\cdot L_v(g)=T_{\pi(g)}\ell_{\tilde g}\cdot \pi_K([g,v+\mathfrak k)], 
$$ 
which is the required equivariancy. 
