Geometric difference between two actions of $GL_n(\mathbb{C})$ on $G\times \mathfrak{g}^*$

Question

Let $G=GL_n(\mathbb{C})$.

Scenerio 1: Let $G$ act on $T^*(G)=G\times \mathfrak{g}^*$ by $$ g.(x,y)=(gx,y). $$

Scenerio 2: Let $G$ act on $T^*(G)=G\times \mathfrak{g}^*$ by $$ g.(x,y)=(xg^{-1},gyg^{-1}). $$

Is there a geometric difference between the two scenerios?

Answer 1

The actions are equivalent. Consider the map $\phi:G\times\frak{g}^*$ $\to G\times\frak{g}^*$ defined by $(x,y)\mapsto (x^{-1},xyx^{-1}),$ where we consider the domain with the first action, and the codomain with the second. Then this map is equivariant, since $$\phi(g\cdot(x,y)) = \phi((gx,y))=(x^{-1}g^{-1},(gx)y(x^{-1}g^{-1}))= g\cdot(x^{-1},xyx^{-1})=g\cdot\phi(x,y).$$

To see that $\phi$ is a bijection is straightforward, thus the $G$-actions are equivalent.