Let $(V,\Omega)$ be a symplectic vector space. In the product $V \times V$, we have the following two symplectic structures: $$(\Omega \ominus \Omega)((x,u),(y,v)) = \Omega(x,y) - \Omega(u,v)$$and $$\Omega_\Omega((x,u),(y,v)) = \Omega(x,v) - \Omega(y,u).$$Is there a symplectomorphism between $(V\times V, \Omega\ominus \Omega)$ and $(V\times V, \Omega_\Omega)$ which does not depend on any choice of basis for $V$? They're so similiar, one might think the answer is "yes". Disentangling the variables seems a bit hard.
Some context for the notation $\Omega_\Omega$. I figured out that if $V$ and $W$ are (finite dimensional) vector spaces and $B:V\times W\to \Bbb R$ is a perfect pairing (i.e., bilinear and the induced maps $V\to W^*$ and $W\to V^*$ are isomorphisms), then $\Omega_B((x,u),(y,v)) = B(x,v)-B(y,u)$ is a symplectic form in $V\times W$. Now, any symplectic form $\Omega$ in $V$ is a perfect pairing of $V$ with itself, so $\Omega_\Omega$ makes sense.
Following Dominique's suggestion in the comments, I'll define $$\begin{align} F\colon V\times V &\to V \times V \\ (x,u) &\mapsto \left(\frac{x+u}{\sqrt{2}}, \frac{x-u}{\sqrt{2}}\right).\end{align}$$Now we check that $F$ is a symplectomorphism$\require{cancel}$. Let's do it:
$$\begin{align} (\Omega \ominus \Omega)&(F(x,u), F(y,v)) = (\Omega\ominus \Omega)\left(\left(\frac{x+u}{\sqrt{2}}, \frac{x-u}{\sqrt{2}}\right),\left(\frac{y+v}{\sqrt{2}}, \frac{y-v}{\sqrt{2}}\right) \right) \\ &= \frac{1}{2} (\Omega \ominus \Omega)((x+u,x-u),(y+v,y-v)) \\ &= \frac{1}{2}(\Omega(x+u,y+v) - \Omega(x-u,y-v)) \\ &= \frac{1}{2}(\cancel{\Omega(x,y)}+\Omega(x,v)+\Omega(u,y)+\cancel{\Omega(u,v)} - \cancel{\Omega(x,y)}+\Omega(x,v)+\Omega(u,y)-\cancel{\Omega(u,v)}) \\ &= \frac{1}{2}(2\Omega(x,v) + 2\Omega(u,y)) \\ &= \Omega(x,v) - \Omega(y,u) \\ &= \Omega_\Omega((x,u),(y,v)),\end{align}$$as wanted.