TITLE: Is there a smooth manifold which admits only rigid metrics?
QUESTION [24 upvotes]: 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), it follows that if $S$ has the above property, then its MCP has only elements of infinite order.

REPLY [5 votes]: The survey paper "Do manifolds have little symmetry?" by Volker Puppe lists several known results about manifolds with no nontrivial finite group action. The ArXiv link is http://arxiv.org/pdf/math/0606714v1.pdf
In particular

aspherical manifolds with $Z(\pi_1M)=0$ and $Out(\pi_1M)$ torsionfree 

have no finite group action (Borel) 
Examples are

certain mapping tori of nilmanifolds (Conner-Raymond-Weinberger)
certain hypertoral manifolds, i.e., n-manifolds with a degree 1 map to the n-torus (Schultz)
certain 3-manifolds (Edwards)
a certain Bieberbach manifold (Waldmüller)