This problem is at the interface of matrix algebra and spectral graph theory.
Let $\mathbf{S}$ be a symmetric $n\times n$ matrix, with positive entries $S_{ij}\geq 0$, and $\mathbf{D} = \mathrm{diag}(\mathbf{S})$ denote the extracted diagonal of $\mathbf{S}$. This would typically be the signless Laplacian of a graph.
Let $\mathcal{E}(\mathbf{S}) = \left\{\mathbf{U} \in \mathcal{SO}(n) \; \left| \;\mathbf{S}' = \mathbf{U}^T \mathbf{SU}, \; S'_{ij} \geq 0, \; \mathbf{D}' = \mathrm{diag}(\mathbf{S}') = \mathbf{D}\right. \right\}.$
In other words, given a symmetric matrix $\mathbf{S}$ with positive entries, $\mathcal{E}(\mathbf{S})$ is the set of special orthogonal matrices $\mathbf{U} \in \mathcal{SO}(n)$ such that the orthogonal transform of $\mathbf{S}$,
$\mathbf{S}' = \mathbf{U}^T \mathbf{SU},$
has all positive elements $S'_{ij} \geq 0$, and has the same diagonal entries as $\mathbf{S}$
$\mathbf{D}' = \mathrm{diag}(\mathbf{S}') = \mathbf{D}$.
Under which condition on $\mathbf{S}$ is the set $\mathcal{E}$ reduced to the singleton identity $\left\{\mathbf{I}_n\right\}$?
This would mean that the matrix $\mathbf{S}$ is entirely determined by any orthogonally similar form (and in particular, the diagonalized form, through the spectral theorem) and its diagonal entries.
Equivalently, under which conditions is a weighted graph entirely determined by the spectrum of its signless Laplacian and the degrees of its nodes?
For example, it would seem to me that having distinct node degrees for a connected weighted graph could be a lead.
Any ideas or related articles are welcome!