Let $A$ and $B$ be two symmetric and positive semidefinite matrices with the same size. Further, assume that $A$ and $B$ share the same column space (i.e., $\mathcal R (A) = \mathcal R (B)$ ).
Is there any relation (even in terms of strong inequalities) between the SVD of $A$ and $B$?
Yes. The number of singular values will be the same and the left singular vectors corresponding to those singular values will match. This is clear since if $A = U\Sigma V^T$ then $$Ax = \sum_{i} u_i (\lambda_i v_i^T x), $$ which shows that the column space of $A$ is the same as the span of the columns of $U$. Since the column space of $A$ and $B$ are the same, the span of their left singular vectors will be the same (and since they have the same range, the rank will be the same). They may be different up to reordering however.
While the result above is true in general, note that since $A$ and $B$ are symmetric positive-semidefinite, this implies that they have the exact same singular value decomposition (except for singular values/eigenvalues). That is $A = U\Sigma U^T$ and the columns of $U$ are the same as the corresponding columns of the eigendecomposition for $B$. Indeed, since they are both psd, their svd's will be the same as their eigendecomposition.