It is well known that the set of fixed points of an isometry $\phi:(M,g)\rightarrow (M,g)$ is a totally geodesic embedded submanifold. (e.g here ).
I ask whether the converse is true, i.e is every totally geodesic embedded submanifold $N \subset M$ can be realized as the set of fixed points of some isometries?
One trivial obstruction is that $N$ should be closed.
(Since the $Id,\phi $ are continuous and the diagonal in $M \times M$ is closed, the set of fixed points is always closed in $M$).
So, if we assume $M$ is connected, then of course we have to omit from our discussions open submanifolds (which I think are all totally geodesic but cannot be closed, hence cannot be a fixed-points-set).
Update: The answer is negative. A brief summary of the idea: take a small enough compact geodesic segement. Any isometry which fixes it, must "fix some more" of the whole geodesic the segment is a part of.
Now I wonder if every such submanifold must be the fixed point of some diffeomorphism(s)? (not necessarily an isometry).
Best Answer
This simply cannot be true. A generic Riemannian manifold does not have any isometries, but still the image of any geodesic is a totally geodesic submanifold.