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Warning 1. I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.

Warning 2. There is an error in my conclusive sentence: the appeal to the method of characteristics is not correct.

Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.

For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.

So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.

Let us introduce the following time dependent forms on $M$:
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.

In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.

Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df$, and determines a unique $X_t$ for each smooth function $f$.

Finally the second condition becomes the following one only on $f$: $Y_t(f)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.

A solution for this equation could be constructed using the method of characterics, considering that $Y_t$ is non singular because such is $dH_t$.

Edit: I was helped to find out, with an explicit counterexample, that my appeal to the method of characteristics is not well grounded. So, in this situation, or we find other conditions, which I did not found, sufficient to apply the method of characteristics, or I need find an alternative strategy.

Warning: I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.

Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.

For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.

So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.

Let us introduce the following time dependent forms on $M$:
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.

In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.

Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df_t$, and determines a unique $X_t$ for each smooth function $f=\{f_t\}_t$ on a neighborhood of $M\times\[0,1\]$.

Finally the second condition becomes the following one only on $f=\{f_t\}_t$: $\mathcal{L}(Y_t).(f_t)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.

A solution for this equation $\mathcal{L}(Y_t+0\frac{\partial}{\partial t}).f=g$ could be constructed using the method of characterics, considering that $Y\equiv Y_t+0\frac{\partial}{\partial t}$ is non singular because such is $dH_t$.

Warning: I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.

Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.

For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.

So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.

Let us introduce the following time dependent forms on $M$:
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.

In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.

Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df$, and determines a unique $X_t$ for each smooth function $f$.

Finally the second condition becomes the following one only on $f$: $Y_t(f)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.

A solution for this equation could be constructed using the method of characterics, considering that $Y_t$ is non singular because such is $dH_t$.

Warning 1. I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.

Warning 2. There is an error in my conclusive sentence: the appeal to the method of characteristics is not correct.

Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.

For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.

So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.

Let us introduce the following time dependent forms on $M$:
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.

In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.

Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df$, and determines a unique $X_t$ for each smooth function $f$.

Finally the second condition becomes the following one only on $f$: $Y_t(f)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.

A solution for this equation could be constructed using the method of characterics, considering that $Y_t$ is non singular because such is $dH_t$.

Edit: I was helped to find out, with an explicit counterexample, that my appeal to the method of characteristics is not well grounded. So, in this situation, or we find other conditions, which I did not found, sufficient to apply the method of characteristics, or I need find an alternative strategy.

Warning: I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.

Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.

For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\phi_{\ast}h_0(x_1)=dh_1(x_1)$.

So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.

Let us introduce the following time dependent forms on $M$:
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.

In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.

Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df$, and determines a unique $X_t$ for each smooth function $f$.

Finally the second condition becomes the following one only on $f$: $Y_t(f)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.

A solution for this equation could be constructed using the method of characterics, considering that $Y_t$ is non singular because such is $dH_t$.

Warning: I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.

Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.

For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.

So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.

Let us introduce the following time dependent forms on $M$:
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.

In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.

Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df$, and determines a unique $X_t$ for each smooth function $f$.

Finally the second condition becomes the following one only on $f$: $Y_t(f)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.

A solution for this equation could be constructed using the method of characterics, considering that $Y_t$ is non singular because such is $dH_t$.