Space of all complex structures on $V$ is homotopy equivalent to the space of (linear) symplectic forms on $V$ How can I prove that the space of all complex structures $I\colon V\to V$ is homotopy equivalent to the space of all non-degenerate skew-symmetric $2$-forms $\omega \in \Lambda^2V$?
Skew-symmetry is seems to be related with the property $I^2=-\operatorname{id}_V$, but I can't formalize it to complete proof.
 A: First of all, recall that a vector space admits a complex structure if and only if it is even-dimensional, which also means it admits symplectic forms.
The proof outlined by Olivier Bégassat in the comments is correct. In this answer, I give some more details but fall short of providing a full proof. As Olivier argued, the idea is to show that the space $\mathcal J(V)\times_\text{comp}\Omega (V)$ of pairs $(J,\omega)$, where $J$ is $\omega$-compatible (meaning $\omega(-,J-)$ is a scalar product) is a fibration over $\mathcal J(V)$ and $\Omega(V)$, the spaces of complex structures and symplectic forms on $V$, respectively, with contractible fibers. This should yield a homotopy equivalence $\mathcal J(V)\simeq \Omega(V)$ after applying the long exact sequence associated to a fibration. In this post, I will focus on the contractibility of the fibers. I invite anyone willing to complete the proof, in which case I will turn this answer into a community wiki.
First, we consider the map $\mathcal J(V)\times_\text{comp} \Omega(V)\to \mathcal J(V)$. The fiber over $J\in \mathcal J(V)$ is the space of symplectic forms that $J$ is compatible with, $\Omega(V,J)$. Note that the fiber over any $J\in \mathcal J(V)$ is not the empty set, since any $J\in \mathcal J(V)$ is compatible with some symplectic form: Using induction, it is not hard to show that every $V^{2n}$ equipped with a complex structure $J$ admits a basis of the form $\{x_1,J x_1,\dots, x_n,J x_n\}$. Declaring this basis to be orthonormal, we obtain a scalar product $g(-,-)$ on $V$ such that $J$ is orthogonal with respect to it. Then $\omega(-,-)=g(J-,-)$ is a symplectic form on $V$ and $J$ is compatible with $\omega$.
It is easy to see that each fiber is a convex space: If $J$ is compatible with $\omega$ and $\omega'$, then for $t\in[0,1]$ the combination $(1-t)\omega(-,J-)+t\omega'(-,J-)$ is an inner product. Thus, the fiber over $J$ is contractible.
Now, consider the fiber of the map $\mathcal J(V)\times_\text{comp} \Omega(V)\to \Omega(V)$, i.e. the space of $\omega$-compatible complex structures $\mathcal J(V,\omega)$. Proving contractibility of this space is more cumbersome, and I will not prove all the details. It is shown in e.g. proposition 2.50 of McDuff and Salamon's book Introduction to Symplectic Geometry. The idea is to show that $\mathcal J(V,\omega)$ is homeomorphic to the space of symmetric, positive-definite symplectic matrices. After picking a standard basis such that $\omega$ is the standard symplectic form on $\Bbb R^{2n}$, this map is given by $J\mapsto J_0J$ where 
$$ J_0=\begin{pmatrix}0 & -\operatorname{id}_n \\ \operatorname{id}_n & 0 \end{pmatrix}$$ represents multiplication by $i$. The inverse map sends $P\mapsto -J_0 P$. To prove contractibility, use the fact that if $P$ is a symmetric, positive-definite and symplectic matrix, then $P^\alpha$ for any $\alpha\in \Bbb R_+$ is also a symplectic matrix, and hence this space is contractible by letting $\alpha$ run from $1$ to $0$, mapping $P$ to the identity matrix.
Now, once we establish that the maps from $\mathcal J(V)\times_\text{comp} \Omega(V)$ are indeed fibrations you obtain the result we were after.
