I thought if there is an operator $\hat A$ which commutes with Hamiltonian $\hat H$, the eigenvalue of the corresponding observable $A$ should be the constant of motion.

In free space $(V=0)$, $$[\hat H, \hat p]=0,\quad \hat p|p\rangle=\mathbf p|p\rangle$$ so we can find that the eigenvalue $\mathbf p$ is the constant of the motion, which is momentum.

In case of crystal, $\hat p$ does not commute with Hamiltonian, but the translation operator with lattice vector $\mathbf R$ commutes with Hamiltonian. $$[\hat H, \hat T(\mathbf R)]=0$$ Also by Bloch's Theorem, there exists quantum number $\mathbf k$ (which looks so similiar with the wave number $\mathbf k$) that satisfies $$\psi_\mathbf k(\mathbf r+\mathbf R)=e^{i\mathbf k\cdot\mathbf R}\psi_\mathbf k(\mathbf r)$$ So I guessed: $$\hat H\psi_\mathbf k(\mathbf r)=E\psi_\mathbf k(\mathbf r), \quad \hat T(\mathbf R)\psi_\mathbf k(\mathbf r)=\psi_\mathbf k(\mathbf r+\mathbf R)=e^{i\mathbf k\cdot\mathbf R}\psi_\mathbf k(\mathbf r)$$ then $\psi_\mathbf k(\mathbf r)$ would be the common eigenstate for both of them, and the eigenvalues would be $E$ and $e^{i\mathbf k\cdot\mathbf R}$.

But I couldn't come up with the observable concept for the translation operator $\hat T$, so I cannot understand why we could use $\mathbf k$ for the constant of motion even if $\mathbf p$ is not.

Does my statement above shows that the steady state solutions are identified with only $\mathbf k$? If so, I think I'm not familiar with the concept of $\mathbf k$. How can I understand $\mathbf k$ intuitively? And how could it be preserved?