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Dear Somnath Basu, the answer to your first question should be yesis that, and for any $k\geq 2$, the k-fold transitivity of the action of $\mathrm{Sympl}(M,\omega)$ on $M$ has only one obstruction, the trivial one, i.e. connectivity of $M$.
But has you proposed there is even more.

Obviously we are assuming connectness otherwise the n-transitiveness is impossible for $n\geq 2$.

In particular Theorem A in a paper of W. Boothby says that,

given a connected symplectic manifold $(M,\omega)$, for any two sets $\{x_1,\ldots,x_n\}$ and $\{y_1,\ldots,y_n\}$ of disjoint point in $M$, where $n$ is arbitrary natural number, there exists a time dependent hamiltonian vector field $X_t$ of $(M,\omega)$ such that its evolution operator $K^X_{1,0}$ maps $x_i$ to $y_i$ for $i=1,\ldots,n$.
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Dear Somnath Basu, the answer to your first question should be yes, and there is even more.

Obviously we are assuming connectness otherwise the n-transitiveness is impossible for $n\geq 2$.

In particular Theorem A in a paper of W. Boothby says that,

given a connected symplectic manifold $(M,\omega)$, for any two sets $\{x_1,\ldots,x_n\}$ and $\{y_1,\ldots,y_n\}$ of disjoint point in $M$, where $n$ is arbitrary natural number, there exists a time dependent hamiltonian vector field $X=\{X_t\} t$ X_t$ of $(M,\omega)$ such that its evolution operator $K^X{1,0}$ K^X_{1,0}$ maps $x_i$ to $y_i$ for $i=1,\ldots,k$.i=1,\ldots,n$.
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Dear Somnath Basu, the answer to your first question should be yes, and there is even more.

Obviously we are assuming connectness otherwise the n-transitiveness is impossible for $n\geq 2$.

In particular Theorem A in a paper of W. Boothby says that,

given a connected symplectic manifold $(M,\omega)$, for any two sets $\{x_1,\ldots,x_n\}$ and $\{y_1,\ldots,y_n\}$ of disjoint point in $M$, where $n$ is arbitrary natural number, there exists a time dependent hamiltonian vector field $ X=\{X_t\} t$ of $(M,\omega)$ such that its evolution operator $K^X{1,0}$ maps $x_i$ to $y_i$ for $i=1,\ldots,k$.