A flow on a manifold $M$ is a one-parameter group of diffeomorphism $\psi_t: M \to M$. In particular, this induces a smooth group action on $M$, $$ \mathbb{R} \times M \to M $$ $$ (t,m) \mapsto \psi_t(m)$$ So that we have a group homomorphism $\phi: \mathbb{R} \to Diff(M)$.
My question is the following: Given any diffeomorphism $g \in Diff(M)$, does there exist a flow (and thus a vector field) on $M$ such that $g$ is in the image of $\phi$?
The answer is no, and the reason may be concisely stated: for any flow
$\phi: \Bbb R \times M \to M \tag 1$
each
$\phi_t: M \to M, \; \forall x \in M \; \phi_t(x) = \phi(t, x) \tag 2$
is homotopic to the identity map, via a homotopy
$H(s, t, x) = \phi_{st}(x), \; s \in [0, 1]; \tag 3$
it is easily seen that
$H(0, t, x) = \phi_0(x) = x, \; H(1, t, x) = \phi_t(x). \tag 4$
For many $M$ there are, however, diffeomorphisms $\psi:M \to M$ which are not homomtopic to the identity; an example is provided by the antipodal map $\psi(x) = -x$ on $S^{2n}$ for any positive integer $n$.
No, the group of diffeomorphism of a manifold is not always connected, for example for the torus, the connected components are in bijection with $SL(2,\mathbb{Z})$.
https://en.m.wikipedia.org/wiki/Mapping_class_group_of_a_surface
