Let $J(V)$ be the set of all complex structures on a finite-dimensional real vector space $V$. (A complex structure on $V$ is by definition a linear isomorphism $J:V\to V$ such that $J^2=-\text{id}$.) Fix such a complex structure $J$, and let $\text{GL}(V,J)$ be the group of all linear isomorphisms $L$ of $V$ such that $L\circ J=J\circ L$. I am trying to show that there is a one-to-one correspondence between the set $J(V)$ and the set $\text{GL}(V)/\text{GL}(V,J)$ of all left cosets.
My attempt: The group $\text{GL}(V)$ acts on $J(V)$ by conjugation, and the stabilizer of $J$ is by definition the subgroup $\text{GL}(V,J)$. But how do we know that this action is transitive?
Here's a sketch: If $J$ is an arbitrary complex structure, there exists an ordered basis of the form $$ (v_{1}, Jv_{1}, v_{2}, Jv_{2}, \dots, v_{n}, Jv_{n}). $$ Conversely, to every ordered basis $$ (v_{1}, w_{1}, v_{2}, w_{2}, \dots, v_{n}, w_{n}) $$ there is a unique complex structure satisfying $w_{k} = Jv_{k}$ for $1 \leq k \leq n$.
The general linear group acts transitively on the set of ordered bases.
Meta-caution: Some authors write "complex bases" in the order $$ (v_{1}, v_{2}, \dots, v_{n}, Jv_{1}, Jv_{2}, \dots, Jv_{n}). $$ This is potentially vexing because a choice of ordered basis determines an orientation on a complex vector space, and these two conventional orderings do not generally define the same orientation.