TITLE: Flows of vector fields and diffeomorphisms isotopic to the identity
QUESTION [22 upvotes]: Let $M$ be a compact manifold and $\varphi : M \longrightarrow M$ be a diffeomorphism which is isotopic to the identity. Does there exist a vector field $ X $ on $M$ such that $\varphi$ is the flow at time $1$ of $X$? If that is not always the case, where does the obstruction for such a $\varphi$ to be a flow lives in ?

REPLY [7 votes]: In fact, for every manifold $M$ with $\dim(M)\ge 2$, the group $Diff_c(M)$ of diffeomorphisms with compact support contains a smooth curve $c:\mathbb R\to Diff_c(M)$
(smooth in the sense that the associated mapping $\hat c: \mathbb R\times M\to M$ is smooth)
with $c(0)=Id$
such that the $\{c(t): t\ne 0\}$ are a free set of generators for a free subgroup of the diffeomorphism group such that no diffeomorphism in this free subgroup with the exception of the identity embeds into a flow.
This is proved in

Grabowski, J., Free subgroups of diffeomorphism groups, Fundam. Math. 131 (1988), 103–121.

The proof uses clever constructions with Nancy Kopell's diffeomorphisms on $S^1$ (those described in Andy Putman answer) which do not embed into a flow.
This free subgroup is contractible to the identity: slide the generators $c(t)$ to $c(0)$.
So the image of exponential mapping of $Diff_c(M)$ does not meet very many diffeomorphisms near the identity.