This question is motivated by the classical fact from differential geometry :

*Let $M$ be a smooth manifold of dimension at least $2$. Then for any $n$ the diffeomorphism group $\textrm{Diff}(M)$ acts transtively on the configuration space of $n$-points in $M$ or equivalently it acts $n$-transitively on $M$.*

As I recall, it is known that the symplectomorphism group $(M,\omega)$ acts transitively on $M$, which is assumed to be symplectic. My question is then the following :

*Let $(M,\omega)$ be a symplectic manifold.
(i) When does $\textrm{Symp}(M,\omega)$ act $n$-transtively for $n\geq 2$ ?
(ii) If the answer above is NOT ALWAYS then what is known ?*

As some background, the usual way one proves (rather the only way I know how to prove this) the first fact is by showing the following :

(i) for two sets of distinct $n$ points in $M$ given by $\{p_1,\ldots,p_n\}$ and $\{q_1,\ldots,q_n\}$ which are close, we find disjoint disks $D_i$'s containing $p_i,q_i$. This requires dimension at least $2$. Use some diffeomorphism of $D_i$ that is smoothly identity at the boundary and looks like a rotation inside $D_i$ that swaps $p_i$ and $q_i$.

(ii) Define the natural equivalence relation on $n$-tuples and observe that the configuration space of $n$-points in $M$. By (ii) each equivalence class is open. It is alsoclosed being the complement of open sets. Since the configuration space is connected (this requires dimension at least $2$) this means there is only one equivalence class.

Does this idea work in the symplectic setting - perhaps by taking paths $\gamma_i$ from $p_i$ to $q_i$ and getting Hamiltonian vector fields via $\omega(\gamma_i',\cdot)$ ?