Let's assume we have a set of 2D-points. My claim is that if that group has at least one valid symmetry axis, then at least one of those axises is equivalent to an eigenvector of the covariance matrix (see en.wikipedia.org/wiki/File:GaussianScatterPCA.png for a more intuitive picture; the black vectors represent the eigenvectors I'm talking about).
For example, let's assume we have points (2,0), (-2,0), (0,1) and (0,-1). The eigenvectors are (1,0) and (0,1) (times some non-zero real number). In this case, those vectors are also the symmetry axises of the set of points.
Then again, take points (0,1), (0,-1), (5,2), (5,-2). Now vector (1,0) is a symmetry axis but (0,1) isn't.
In both cases at least one eigenvector is also a symmetry axis. Is this always the case?
Let $R$ be the orthogonal matrix representing reflection along your symmetry axis. The assumption that the point set is symmetric then implies that $RMR^t=M$ or said differently, $RM=MR$. Now $R$ has the form $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ for suitable choice of basis and all the matrices commuting with that matrix are of the form $\begin{pmatrix}r&0\\0&s\end{pmatrix}$ which shows that the reflection line and its orthogonal complement are eigenvectors of $M$.
EDIT: Sorry I misread the question. I thought you asked whether any symmetry axis is also an eigenvector. The converse is of course not true. Just pick a finite set of points without any symmetry axis,e.g. the vertices of the unit sqaure and $(2,3)$. Then the covariance matrix is still a symmetric matrix and thus is is diagonalizable, but there are no symmetry axes.
EDIT2: If I scrolled down in the comments, I would have seen that my answer is the same as Anton Malyshev's comment. I would delete it, but this would probably mean that this question stays open.