Consider a compact differentiable manifold $M$. We say that $f:M\to M$ and $g: M \to M$ are topologically conjugated if there exists $h:M\to M$ a homeomorphism such that $f\circ h= h \circ g$. The conjugacy class of a homeomorphism $f$ is the set of all $g$ such that $g$ is topologically conjugated to $f$.


If a homeomorphism $f: M\to M$ has infinite topological entropy (which is an invariant under topological conjugacy), then the conjugacy class of $f$ has no diffeomorphisms.


Is there any other known obstruction for a homeomorphism not to have diffeomorphisms in its conjugacy class? I would guess that yes, but I could not find one. 


Is there a restriction on the dimension of the manifold? 


(Remark: In dimension one, that is, in the circle, every homeomorphism has a diffeomorphism-of class $C^1$ in its conjugacy class. However, Denjoy counterexamples have no $C^2$ diffeomorphisms in theirs). 



Topological conjugacy between homeomorphisms and diffeomorphisms
Here is an example. Consider a map $f:\mathbb R^2\to\mathbb R^2$ given by $(r,\varphi)\mapsto (r,\varphi+\sin(1/r))$ in polar coordinates $(r,\varphi)$, $0<r\le 1/\pi$. For $r>1/\pi$, let $f$ be identity, then close up the plane to make a compact manifold.


This map has zero topological entropy but has no conjugate $C^1$ diffeomorphism. Indeed, we may assume that 0 is mapped to itself by the conjugation. Then the circles centered at the origin are mapped to Jordan curves winding around the origin. There are arbitrarily small circles made of fixed points, hence the derivative of the diffeomorphism at the origin is the identity. On the other hand, there are arbitrary small circles on which our map is periodic with rotation number, say, 1/10. It is easy to see that this cannot happen on a Jordan curve winding around the origin, unless some point changes is angular coordinate by $2\pi/10$. This contradicts the fact that the derivative at 0 is the identity.


