Density operator in different bases This may sound like a silly question, and it probably arises from a weak understanding of the concept.
What I was wondering is: given a density operator $\varrho=\sum_{i,j}p_{ij}|i\rangle\langle j|$, can we find separate bases for which $\varrho$ is diagonal, but presenting different eigenvalues?
Mainly I was considering the example of when $\dot{\varrho}=0$. In this case $\varrho$ is diagonal in the Energy eigenbases: this grants us a nice representation of $\varrho$ as we can clearly observe its eigenvalues for each energy level. My question would then be: is possible to find a second bases(not necesseraly an observable eigenbasis) for which $\varrho$ becomes diagonal but presenting different entries?
Intuitively I would think that this should obviously possible as long as some commutation relation is respected ($=0$), but the whole idea is becoming quite confusing to me.
 A: If I understand your question correctly, then the answer is no.  The eigenvalues of an operator - which are simply the diagonal entries when that operator is diagonalized - are uniquely defined up to trivial reordering. Furthermore, the vectors belonging to distinct eigenspaces are orthogonal because $\rho$ is self-adjoint and therefore symmetric.

so if I am given a diagonal density operator, and then someone claims that ϱ˙=0, then I know for a fact that the ϱ I was given is being represented in the energy basis, right?

No, not quite.  If two energy eigenspaces appear in the density operator with the same probability, then you might use a basis in which $\varrho$ is diagonal but $H$ is not.
For example, consider the 2-level Hamiltonian and density operator in some basis $\{|1\rangle,|2\rangle\}$. $$H=\pmatrix{E&0\\0&-E} \qquad \varrho = \pmatrix{0.5 & 0 \\ 0 & 0.5}$$
Obviously $[\varrho,H]=0 \implies \dot\varrho = 0$.  However, if we change to a new basis
$$|1\rangle = \frac{|a\rangle+|b\rangle}{\sqrt{2}} \qquad |2\rangle = \frac{|a\rangle-|b\rangle}{\sqrt{2}}$$
then the Hamiltonian and density operator become
$$H = \pmatrix{0 & E \\ E & 0}\qquad \varrho = \pmatrix{0.5 & 0\\ 0 & 0.5}$$
As a result, even though $\dot \varrho = 0$, we have a basis $\{|a\rangle,|b\rangle\}$ which is not an energy eigenbasis, but in which $\varrho$ is diagonal.
This is a nice reminder that $[\varrho,H]=0$ implies that there exists a basis in which $\varrho$ and $H$ are simultaneously diagonalized; however, this does not imply that every basis which diagonalizes one also diagonalizes the other.
