Suppose $M$ is a smooth manifold and $x,y \in M$ are two points. Is there always a
diffeomorphism $\phi: M \rightarrow M$ with $\phi(x)= y$ ?



Automorphism of Smooth manifolds
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. 


