High multiplicity eigenvalue implies symmetry? It is well known that on any compact Riemannian symmetric space $X$, the eigenvalues of the Laplacian have very high multiplicity (comparable with the Weyl bound), and the resulting actions $\operatorname{Isom}(X)\to \operatorname{SO}(W_\lambda)$ for eigenspaces $W_\lambda$ give many representations of the Lie group $\operatorname{Isom}(X)$.
Suppose one has an unknown compact Riemannian manifold $X$ ($n=\dim X$), but where the eigenspaces of the Laplacian have large dimension (I don't have a precise definition of "large" here; the weakest definition would probably be something like $\dim W_\lambda>1$ for infinitely many $\lambda$.  I'd be happy even with a much stronger assumption, say $\dim W_\lambda$ is at least $\epsilon$ times the Weyl bound $\operatorname{const}\cdot\lambda^{(n-1)/2}$ infinitely often).  Can one conclude that $X$ is a symmetric space, or close to one in some sense?
EDIT: I would be interested in any result which takes as a hypothesis some assumption of large multiplicity in the Laplace spectrum, and whose conclusion is some sort of symmetry of the underlying manifold.
 A: One situation where people have thought hard about this issue is the multiplicity of the spectrum of the Laplacian on the modular surface. This is a notoriously difficult problem. 
Conjecturally, the (discrete) spectrum is simple, but as far as I know, the best known bounds  on the multiplicity of a Maas cusp form of eigenvalue $\lambda$ are somewhere in the neighborhood of  $O(\lambda^{1/2}/\log \lambda).$
(Weyl's law in dimension $2$ says that the number of eigenvalues of size at most $N$ is $O(N)$). 
See e.g. section 4 of the following survey of Peter Sarnak: http://www.math.princeton.edu/sarnak/baltimore.pdf
So I would say that some of the stronger versions of your conjecture are almost certainly out of reach. 
A: My guess is No. You do not need a one parameter Lie group of symmetry to have infinitely many double eigenvalues. Just one involution suffices. And one involution is not enough to make a symmetric space. For instance, every complex curve whose equation is real has this property; then the involution is the complex conjugation. For most such curve, there is no other non-trivial involution.
A: I might be totally confused, but in this paper of Hubert Goldschmidt (Infinitesimal isospectral deformations of symmetric spaces), he seems to construct the very deformations of the title. It is not obvious that these can be integrated, but if they can be, it gives you a family of manifolds isospectral to the symmetric spaces and not symmetric, which would provide a negative answer to the OP (note that if you read Goldschmidt's paper, you will see that various spaces are not NOT to be so deformable).
A: My answer is rather a question. Suppose that $(M,g)$  is a compact Riemann manifold and for any  positive integer $N$ there exists an eigenvalue of the Laplacian that has multiplicity $>N$. Can one conclude that the group of isometries of $(M,g)$ has positive dimension?
