OP asks good questions.
First of all, let us mention that there is a bijective correspondence between 1-parameter symplectic flows and parameter-independent$^1$ symplectic vector fields. The latter are by definition vector fields $X\in \Gamma(TM)$ that preserve the symplectic 2-form ${\cal L}_X\omega=0$.
OP is essentially asking the following (some of it in previous versions of the question).
- Is the symplectomorphism group path-connected?
Answer: Not necessarily, there could be topological obstructions.
2D counterexample: Let the phase space $M=\mathbb{S}^2$ be the 2-sphere equipped with the standard symplectic 2-form $\omega$. The second homotopy group $\pi_2(\mathbb{S}^2)\cong\mathbb{Z}$ is non-trivial. $\Box$
- Is any symplectomorphism the 1-time of a 1-parameter symplectic flow, i.e. is it the exponential $\exp(X)$ of a symplectic vector field $X$?
Answer: Not necessarily, there could be topological obstructions.
Conjectured 2D counterexample: Consider the phase space $M=\mathbb{R}^2$ with the symplectic 2-form $\omega =\mathrm{d}p\wedge \mathrm{d}q$. Consider the canonical transformation
$$ \begin{bmatrix} Q \cr P \end{bmatrix}
~=~ A\begin{bmatrix} q \cr p \end{bmatrix}, \qquad
A~:=~\begin{bmatrix} -2 & 1\cr 3 & -2 \end{bmatrix}~\in~Sp(2,\mathbb{R})~=~SL(2,\mathbb{R}). $$
We claim that this symplectomorphism can not be generated by a symplectic flow. A tell-tale fact is that the matrix $A$ has no square root.
Another clue is that this symplectomorphism only has the origin as a fixed point. This means that a symplectic vector field $X$ (for a flow, if it exists) can at most vanish at the origin.
Interestingly, one can write this symplectomorphism as a canonical transformation with a type-2 generating function
$$F_2(q,P)~=~-\frac{1}{2}qP +\frac{3}{4}q^2 - \frac{1}{4}P^2. $$
However we conjecture that a 1-parameter deformation from the identity $F_2(q,P)=qP$ must always go through a singular point.
See also this related Math.SE post. $\Box$
- Is any symplectic vector field $X\in \Gamma(TM)$ a Hamiltonian vector field?
Answer: Not necessarily, there could be topological obstructions. In fact, this is measured by the first Poisson cohomology group.
2D counterexample: Consider the phase space $M=\mathbb{R}^2\backslash\{(0,0)\}$ with the symplectic 2-form $\omega =\mathrm{d}p\wedge \mathrm{d}q$. One may check that the vector field
$$X=\frac{q}{q^2+p^2}\frac{\partial}{\partial q} +\frac{p}{q^2+p^2}\frac{\partial}{\partial p} $$
is symplectic but it is not a Hamiltonian vector field. The problem is that the candidate ${\rm arg}(q+ip)$ for the Hamiltonian generator is multi-valued, and hence not globally well-defined.
See also e.g. this & this related Phys.SE posts. $\Box$
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$^1$ Note that there exists a notion of flow corresponding to a parameter-dependent vector field. We shall not consider this in this answer. Such flows are used in e.g. this related MO.SE post.
