Isometry Group of a Manifold Let $(M,g)$ be a Riemannian manifold and let $I = Iso(M)$ be the group of isometries of $M$. Suppose that $I$ has no subgroups. What does this tell us about $M$?
 A: Here are some details for my comments.
Claim. Suppose that $M$ is a smooth manifold of dimension $\ge 2$. Then it admits a Riemannian metric with no symmetries. 
Suppose that $M$ has dimension $\ge 3$. Then it is proven in 
J. Kazdan, F. Warner, "Scalar curvature and conformal deformation of Riemannian structure," J. Differential Geometry 10 (1975), 113–134 
that every $C^\infty$-function $K: M\to {\mathbb R}$ which is negative at some point, equals the scalar curvature  of a $C^\infty$-Riemannian metric on $M$. 
Since isometries of a manifold have to preserve scalar curvature, we can just take a "random" smooth function $K$ on $M$: It will have no symmetries and, hence, the corresponding Riemannian metric will have no symmetries either. 
For 2-dimensional manifolds there are more restrictions on scalar/Gaussian curvature. However, one can use the results of 
J. Kazdan, F. Warner, "Curvature functions for open 2-manifolds", Ann. of Math. (2) 99 (1974), 203–219
and 
J. Kazdan, F. Warner, "Curvature functions for compact 2-manifolds," Ann. of Math. (2) 99 (1974), 14–47. 
to get the same conclusion. 
In the 1-dimensional case (I am assuming the manifold is connected), if the Riemannian manifold is compact, then it is isometric to a circle of certain radius and, hence, has 1-dimensional group of symmetries. For noncompact manifolds, one can take the metric isometric to the half-line; such metric has no symmetries. 
