In a real finite vector space $V$ with a non-degenerated skew-symmetric form $\omega: V \to V$ there is always a so called symplectic basis $\{x_1,y_1, \cdots, x_n,x_n\},$ according to which $\omega$ is written as $$\omega = \sum_i dx^i\wedge dy^i,$$ or in other words, a basis such that $\omega(x_i,y_i) = 1$ for every $i$ and any other pairing vanishes. This is equivalent to saying that the matrix associated to $\omega$ in this basis has $1$'s above the diagonal, $-1$'s under it and zero everywhere else.
Now, if I have other non-degenerated skew-symmetric form $\widetilde\omega$, could I possibly find a basis for $V$ where the matrices of both $\omega$ and $\widetilde\omega$ are as described above, but with constants different than one? What I mean is, are there conditions for the skew symmetric forms that guarantee the existence of a basis $\{x_1,y_1, \cdots, x_n,x_n\}$ with respect to which one has $$\omega = \sum_i \alpha_idx^i\wedge dy^i$$ $$\widetilde\omega = \sum_i \widetilde\alpha_idx^i\wedge dy^i,$$ so that they are simultaneously written on a "canonical" simpler way?
The usual construction of a symplectic basis for a skew-symmetric form doesn't seem to apply in this situation because despite being able to construct a two-dimensional subspace $W \subset V$ that is symplectic with respect to both $\omega$ and $\widetilde\omega$ we have no control over what happen to the intersection of their orthogonal complements $W^\omega$ and $W^{\widetilde\omega}$ in order to try an induction argument, which makes me think such a basis should not always exist.
I'm looking into the case where there are two forms involved to get an idea of what can happen, but I'm actually more interested in the situation where one has a finite number $k$ of skew-symmetric forms involved.