Automorphism of Smooth manifolds Suppose $M$ is a smooth manifold and $x,y \in M$ are two points. Is there always a
diffeomorphism $\phi: M \rightarrow M$ with $\phi(x)= y$ ?
 A: Here's another argument, exploiting the differentiable structure a bit more (in the sense that I don't see a straightforward way to adapt it to the topological/PL case). It might be the standard argument to show that things work locally, but it works globally as well.
Take a smooth, embedded path $\gamma: [0,1]\to M$ so that $\gamma(0)=p$ and $\gamma(1)=q$.  Since $\gamma$ is embedded, we can push forward $\partial/\partial t$ and get a vector field $W$ on the image of $\gamma$. Let $V$ be a compactly supported extension of $W$ to all of $M$.
Then it's easy to check that the flow of $V$ at time 1 is a diffeomorphism of $M$ that sends $p$ to $q$ (and is isotopic to the identity).
This argument readily generalises to $k$-transitivity, too: if $\dim M\ge 3$, we can choose generic paths $\gamma_i$ simultaneously, and use the same exact argument. If $M$ is a surface, we need to use the fact that a single embedded path doesn't disconnect $M$.
A: Partially this is a response to Mariano's 2nd comment. 
In the smooth manifold case there's actually a really slick proof.  Here it is:
Let $\gamma : [0,1] \to M$ be a smooth path in $M$ such that $\gamma(0)=p$ and $\gamma(1)=q$.  You can re-consider this map to be an isotopy from the $0$-dimensional submanifold $\{p\}$ to the submanifold $\{q\}$.  
The isotopy extension theorem then says, there exists a smooth map 
$$G : [0,1] \times M \to M$$
such that the function $G_t : M \to M$ given by $G_t(x) = G(t,x)$ is a diffeomorphism for all $t \in [0,1]$, and $G_0 = Id_M$.  Also, it guarantees $G(t,p) = \gamma(t)$.  So $G_1 : M \to M$ is a diffeomorphism that sends $p$ to $q$. 
The nice thing about this argument is it readily generalizes.  For example, take the configuration space of $k$ distinct points in the manifold $M$, $C_k M$.   
$$C_k M = \{(p_1,\cdots,p_k) \in M^k : p_i \neq p_j \forall i \neq j\}$$
It's not too hard to argue that if $dim(M) \geq 2$ and $M$ is connected, then $C_k M$ is connected.  So isotopy extension kicks in again and says, if $(p_1, \cdots, p_k)$ are $k$ distinct points in $M$ and $(q_1, \cdots, q_k)$ are also $k$ distinct points, then there exists a diffeomorphism $f : M \to M$ such that $f(p_i) = q_i$ for all $i$.    Another way to say this is that $Diff(M)$ acts $k$-transitively on the manifold. 
For $1$-dimensional manifolds, generally $C_k M$ is not connected, and $Diff(M)$ is not $k$-transitive. 
