Cutlocus and conjugate points I am thinking about the following questions about the cutlocus of a point in a Riemannian manifold or of a hypersurface in the Euclidean space: 
1) If all the points of the (nonvoid) cutlocus of a point $p$ in a Riemannian manifold are also conjugate points, then the manifold is rotationally symmetric around the point $p$ and the cutlocus is a single point?  
2) If all the points of the (nonvoid) cutlocus of a smooth, connected hypersurface embedded in ${\mathbb R}^{n+1}$ are also conjugate points, what such a hypersurface looks like?
One guess is that it should be a level set of the distance function from the cutlocus itself. In the case $n=1$, that is, curves in the plane, it is a circle.
It can be seen that in this case, by Sard theorem, the cutlocus has Hausdorff $H^n$-measure zero.
 A: It is known that for a simply-connected compact 
Riemannian symmetric space, the cut locus of a point
coincides with its first conjugate locus, see e.g. the book by Cheeger and Ebin,
"Comparison theorems in Riemannian geometry", Theorem 5.13. So the answer to (1) is no.
Examples of compact symmetric spaces include spheres, projective spaces (real, complex,
quaternionic, octonionic), Grassmannian manifolds, Lie groups with bi-invariant metrics etc.
So you see that this class of Riemannian manifolds is not small.
Further, the cut locus of a point in a symmetric space can be explictly described, and it is in general not a point. In fact, it is a highly non-trivial result, proved by Berger 
(Green in the $2$-dimensional case), that a manifold with the property that the cut locus of any point is reduced to a single point ("wiedersehens manifold") has to be isometric to a round sphere. 
A: For question 1, the answer is NO.
Take any rotationally symmetric manifold with fixed point $p$. 
Perturb it smoothly it in the directions tangent to the spheres centered at $p$. 
You still get one point as the cutlocus of $p$.
