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Neat solutions to tedious questions

Let $M$ be a smooth manifold, $x,y\in M$. Must there exist a diffeomorphism $f : M \rightarrow M$ with $f(x) = y$?

I tried proving this via vector fields, i.e. trying to find a vector field whose flow through $x$ passes through $y$, without much success. Besides, this only has a chance of working on complete manifolds. Anyone know the answer to this?

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No; take $M$ to be the disjoint union of two smooth manifolds which are not diffeomorphic.

However, the statement is true if $M$ is connected. You do not need completeness. It suffices to show that the set of all points that can be reached from $x$ via some diffeomorphism is both open and closed.

You can find a demonstration of this fact (if M is connected) in the book of Milnor – Topology from the differentiable viewpoint. It is the lemma of homogeneity. In fact you have more :

**Homogeneity Lemma: Let $y$ and $z$ be arbitray interior points of the smooth, connected manifold M. Then there exists a diffeomorphism $f:M\rightarrow M$ that is smoothly isotopic to the identity and carries $y$ into $z$.**

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