Proving that the smooth automorphism group of a manifold $M$ acts transitively on $M$ Let $M$ be a differentiable manifold of dimension $n$, let $p,q\in M$ any two points. We need to show there exists an automorphism $f\in \mathrm{Diff}(M)$ with the property that $f(p)=q$.
Could anyone help me to start or any reference to read?
 A: The optimal strategy depends on the tools you already have at your disposal. If you know about flow maps of vector fields, you can just define a field that flows from $p$ to $q$ along some curve, extend it to $M$ continuously (in any way you wish), and flow with it. 
Added: Take a smooth curve $\gamma$ going from $p$ to $q$ with unit speed. Its velocity vector $\gamma'$ defines a vector
field on the image of $\gamma$. You can extend it continuously by using a partition of unity $(U_j,\phi_j)$ subordinate to charts $F_j: U_j\to \mathbb R^n$ (let's say $F_j$ is onto). Only the charts that overlap the image of $\gamma$ should
be considered. For each $j$ the problem is moved to $\mathbb R^n$ where it reduces to extending a vector-valued continuous function defined on a closed set; this is possible by Tietze's theorem. Then move the extended field back to $U_j$ (diffeomorphisms allow you to push and pull, in both directions). Finally, take the sum of such extensions multiplied by $\phi_j$. The result is a smooth vector field on $M$ (actually, zero away from the curve) which coincides
with $\gamma'$ on the image of the curve. Therefore, the flow under this field will arrive from $p$ to $q$ in finite 
time (equal to the length of $\gamma$).

A bare-hands approach may look like this: connect $p$ to $q$ by a smooth curve, take a tubular neighborhood of this curve, observe it's a topological ball, then show (by an explicit formula on   $B=\{x\in\mathbb R^n:|x|<1\}$) that it is possible to map one point of $B$ to any  other while keeping everything near the boundary fixed.  
