Symplectomorphism from one point to another in a connected symplectic manifold I was taking a look at the following old question and I was wondering if this could be extended to symplectic manifolds. So if we have a symplectic manifold $(M,\omega)$ and two different points $x,y \in M$, does there exist a symplectomorphism $\phi: M \to M$ such that $\phi(x) = y$?
I tried the same approach as in that question but I couldn't prove that $\tilde{X}$ was a symplectic vector field.
 A: Define the following equivalence relation on $M$: $x \sim y$ if there exists a symplectomorphism connecting $x$ and $y$. It is easily checked to be an equivalence relation, and it now suffices to prove that each class is open (since $M$ is connected, then there is only one class and the result follows).
So, let $x \in M$, and pick a Darboux chart  $(\phi,U)$. Let $y \in U$. 
The local constant vector field $V \equiv \phi(y)-\phi(x)$ yields a symplectic vector field which has its flow at time $1$ connecting $\phi(x)$ and $\phi(y)$. Since we are on $\mathbb{R}^{2n}$, $\iota_V \omega=dH$ for some $H$ (actually, furnishing such an $H$ manually shouldn't be hard). 
Pick a bump function $\rho$ which is $1$ on an open set containing $x$ and $y$, and its support is contained in $U$. Then $\lambda:=\rho \cdot (H \circ \phi^{-1}) $ is a smooth function for which the Hamiltonian vector field yields a flow such that the time $1$ takes $x$ to $y$. Since $y \in U$ is arbitrary, we have our result.
A: Hint: the question is local, you can use Hamiltonian vector fields. You can suppose that a chart of $M$ is an open subset of $\mathbb{R}^n$ with the restriction of the standard symplectic form (Darboux), consider a function $f$ with compact support which is affine in $V\subset U$ and which vanishes outside of $V$. Let $X_f$ be the Hamiltonian of $f$ shows that by using local coordinates that if $x,y$ are in $V$, there exists an $f$ like above whose flow enables you to connect $x$ and $y$.
