Nash isometric embedding theorem with keeping the symplectic structures of our ambient spaces I apologize in advance if this question has an obvious answer.
Let $(M,g)$ be a Riemannian manifold.
Then the tangent bundle $TM$ carries a natural symplectic structure $\omega_g$. In fact  $\omega_g$ is  the  pull  back of the  canonical  symplectic  structure of the  cotangent bundle  via  the  obvious  diffeomorphism  between $TM$  and  $T^* M$ which is  defined by the inner product $g$.
The standard structure of $T\mathbb{R}^n=\mathbb{R}^n \times \mathbb{R}^n$ is denoted by $\omega$.

For every Riemannian manifold $(M,g)$, is there an isometric embedding $j$ of $M$ in some $\mathbb{R}^n$ such that $j^*(\omega)=\omega_g?$

 A: This is always the case (using "naturality" as Paul Bryan suggested in the comments).
Let $f: M \to N$ be a smooth map between Riemannian manifolds $(M,g)$ and $(N, h)$. Let $g^\flat: TM \to T^*M$ denote the musical isomorphism induced by $g$ (and similarly $h^\flat$).
Now, $f$ is an isometry if and only if $h_{f(m)} (T_m f (X), T_m f (\cdot)) = g_m (X, \cdot)$ holds for all $X \in T_m M$. In other words, $g^\flat (X) = f^* (h^\flat (T_m f(X)))$ or shorter $$g^\flat = T^* f \circ h^\flat \circ Tf,$$ where $T^*f: T^* N \to T^* M$ is the cotangent lift of $f$ (often called a point transformation).
Using the fact that cotangent lifts are always symplectic maps, we have
$$\omega_g = (g^\flat)^* \omega_M = (Tf)^* (h^\flat)^* \omega_N = (Tf)^* \omega_h,$$
where $\omega_M$ and $\omega_N$ denote the canonical symplectic forms on $T^* M$ and $T^* N$, respectively.
A: 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.  
