Let $E$ be a vector space. Let's consider the action group of $GL(E)$ on $E$, defined as: $$\forall g \in GL(E), \forall v \in E, g \cdot v = g(v)$$ What is the number of orbites for this action?
So far I just tried to understand better this action. If $x,y \in E$ and $x \sim y$ it means that $\exists g \in GL(E)$ such that $g(x) = y$. Now I don't know how to go a step further and see how many orbits will there exist. As far as I think, and for any $x$ and $y$ in $E$, there will always exist $g \in GL(E)$ such that $g(x)=y$, thus for any $(x,y)$ we can find a $g \in GL(E)$ such that $g(x)=y$, thus the whole set $E$ can be put in relation with any $x \in E$. Thus there would exist only one orbit.
Is my thinking correct?
There are precisely two orbits of the standard action of $\text{GL}(n,K)$ on $K^n$. The group acts transitively on $K^n -\{(0,\dots,0)\}$, and the zero vector is a fixed point of the action.
Fun fact: One can see $\text{GL}(n,K)$ as an algebraic group, in which case one of these orbits is open, and the other is closed (with respect to the Zariski topology).