Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?


Of course, such a manifold must not admit a diffeomorphism of finite order.


Since a surface $S$ admits a diffeomorphism of order $n$ iff its mapping class group (MCP) has an element of order $n$ (see [here](https://mathoverflow.net/a/30445/46290)), it follows that if $S$ has the above property, then its MCP has only elements of infinite order. 



Is there a smooth manifold which admits only rigid metrics?
The answer to the question in the first sentence is "yes". Let $M$ be a hyperbolic 3-manifold whose isometry group is trivial. Then by Theorem 1.1 of


Farb, Benson; Weinberger, Shmuel Hidden symmetries and arithmetic manifolds. Geometry, spectral theory, groups, and dynamics, 111–119, Contemp. Math., 387, Amer. Math. Soc., Providence, RI, 2005. (Reviewer: Bachir Bekka) 53C35 (53C23)


(which they attribute to Borel, though they do not give an original reference for it), the isometry group of every Riemannian metric on $M$ is isomorphic to a subgroup of the hyperbolic isometry group, and thus is trivial.


Along similar lines, you might be interested in Theorem H of 


Dinkelbach, Jonathan and Leeb, Bernhard,
Equivariant Ricci flow with surgery and applications to finite group actions on geometric 3-manifolds.
Geom. Topol. 13 (2009), no. 2, 1129–1173. 


It says that every finite-order diffeomorphism of a closed hyperbolic 3-manifold is smoothly conjugate to an isometry, so closed hyperbolic 3-manifolds with trivial isometry groups give examples of smooth manifolds with torsion-free diffeomorphism groups.


