It seems here we have an answer: https://www.sciencedirect.com/science/article/pii/S0926224502000670

This is how I would approach the question:

The question is equivalent to the existence of some immersion $f : M \to \mathbb{R}^N$ such that if $\omega_{\mathbb{R}}$ is the sympletic form of $\mathbb{R}^N$ and $\omega$ is the sympletic form of $M$, one has $f^{\ast}(\omega_{\mathbb{R}}) = \omega.$

Note that since $d\omega_{\mathbb{R}} = 0$ and $\mathbb{R}^N$ is simply connected, there exists $\theta \in \Omega^1(\mathbb{R}^N)$ such that $\omega_{\mathbb{R}} = d\theta.$ Therefore,
$$f^{\ast}(d\theta) = \omega.$$ Therefore, $d(f^{\ast}\theta) = \omega.$ This implies a necessary condition is that $\omega$ is exact. If $H_{dR}^2(M) = 0$, then $\omega = d\tilde \theta$, for $\tilde \theta \in \Omega^1(M)$. This implies that:

$$d(f^*(\theta)) = d\tilde \theta.$$ Therefore,
$$f^{\ast}(\theta) - \tilde \theta \in H_{dR}^1(M).$$ If we assume that this is zero, $f^{\ast}{\theta} = \tilde \theta$. Therefore,
$$\tilde \theta (X) = \theta(df(X)), ~\forall X \in TM.$$ Note that $\theta = \langle Z,\cdot\rangle,$ for some vector field $Z \in T\mathbb{R}^N$. Therefore,
$$\theta(X) = \langle df(X),Z\rangle,$$
and this equations suggests an equation for the isommetric immersion.

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