Let $G$ be a diffeomorphism of $\mathbb{R}$. Then the map $\mathbb{R}\times\{0\}\rightarrow\mathbb{R}\times\{0\}$ given by $(x,0)\mapsto (G(x),0)$ may be extended to a symplectomorphism $(x,\xi)\mapsto (G(x),(G')^{-1}(x)\xi)$ of $T^*\mathbb{R}$. (Here, $(x,\xi)$ are symplectic coordinates in $T^*\mathbb{R}$.)

In other words, a map on the zero section of $T^*\mathbb{R}$, of the form given above (i.e., which is identifiable with a diffeomorphism of $\mathbb{R}_x$), may be lifted to a symplectomorphism of the cotangent bundle. (This could certainly be stated in higher dimensions, with $G'$ replaced by the Jacobian matrix of $G$.)

Now let $\mathcal{C} = \{\xi_1 = 0\}\subset T^*\mathbb{R}^2$. We take $(x_1,\xi_1;x_2,\xi_2)$ as coordinates on $T^*\mathbb{R}^2$. Suppose we start with the map $\Psi:\mathcal{C}\rightarrow\mathcal{C}$ given by $(x_1,0;x_2,\xi_2)\mapsto (\varphi(x_1,x_2,\xi_2),0;x_2,\xi_2)$. Here, $\varphi$ is differentiable in all three components and for each $(x_2,\xi_2)$, $x_1\mapsto\varphi(x_1,x_2,\xi_2)$ is invertible. $\Psi$ is a pre-symplectomorphism of the coisotropic submanifold $\mathcal{C}$, with respect to the form $d\xi_2\wedge dx_2$. Think of $\varphi$ as a diffeomorphism of $x_1$ depending smoothly on $(x_2,\xi_2)$: $\varphi(x_1,x_2,\xi_2) = G_{(x_2,\xi_2)}(x_1)$. My question is: Can $\Psi$ be extended to a symplectomorphism of $T^*\mathbb{R}^2$? That is: Does there exist a symplectomorphism $\Phi$ of $T^*\mathbb{R}^2$ such that $\Phi|_\mathcal{C} = \Psi$?

A first attempt would be $\Phi:(x_1,\xi_1;x_2,\xi_2)\mapsto (G_{(x_2,\xi_2)}(x_1),(G'_{(x_2,\xi_2)})^{-1}(x_1)\xi_1;x_2,\xi_2)$, where the prime indicates differentiation in the $x_1$ variable. But this isn't a symplectomorphism.

Any help would be greatly appreciated!