Assume that $M$ is a compact simply connected Riemannian manifold and $f$ is a conformal diffeomeorphism of $M$. Is it true to say that $f$ is homotopic (or isotopic) to an isometry of $M$?
What would be the answer if we drop the simply connected assumption?
On
It is known that if the group of conformal transformation of a compact manifold is compact, then it is conjugate to a group of isometries, and if it is not compact, then the manifolds is the sphere with its standard conformal structure (Lelong-Ferrand Obata).
Assume that the manifold is not conformal to the sphere, then any conformal map is an isometry of another metric conformal to the first one. But perhaps the manifod has no isometry at all ; for instance it is easy to construct a metric on the 2-sphere with no isometry, but this metric has a lot of conformal maps.
In the case the manifold is (conformal to) the sphere, the group of conformal transformations is the Möbius group which contains the groups of isometries as a retract, so every conformal transformation is isotopic to an isometry.
I don't think that this can be true in the form you state it. The point is that you can replace the given metric on $M$ by any conformal rescaling without changing the the conformal diffeomorphisms. However, a generic conformal rescaling of a given metric on $M$ should not admit any isometries. So you would end up with the statement that any conformal diffeomorphism is homotopic to the identity.