Suppose we don't know (yet) about the angular momentum operator, and just associate a (unitary) rotation operator $\hat{R}$ to every element $R$ of the 3D rotation group.

My question is : why does the position $\hat{\boldsymbol{X}}$ operator transforms like a regular vector when the system is rotated : $$ \langle \hat{X}_i \rangle_{\Psi'} = R_{i,j} \langle \hat{X}_j \rangle_{\Psi} \quad \text{with} \quad |\Psi'\rangle=\hat{R}|\Psi\rangle$$ $R_{i,j}$ being the rotation matrix.

Sure, we can invoke the correspondence principle, but I was given an argument the following argument based on galilean relativity : the wave function must transform like $$ |\Psi'(\mathbf{x})|=|\Psi(R^{-1}\mathbf{x})| $$ so that the probabilities are the same after the transformation. Hence, $$ \Psi'(\mathbf{x}) = \mathrm{e}^{\mathrm{i}\alpha(\mathbf{x})}\Psi(R^{-1}\mathbf{x}) $$ with an unknown phase $\alpha(\mathbf{x})$. Then, we can show that for the angular momentum $\langle \hat{\boldsymbol{L}} \rangle$ to stay the same (galilean relativity), we must have a constant phase, which can be eliminated : $$ \Psi'(\mathbf{x}) = \Psi(R^{-1}\mathbf{x}) $$ From that, the vector-like behavior of $\hat{\boldsymbol{X}}$ follows.

While I find this argument convincing, I would like to know if there are more fundamental arguments/postulates without imposing $\langle \hat{\boldsymbol{L}} \rangle$.

Suppose we don't know (yet) about the angular momentum operator, and just associate a (unitary) rotation operator $\hat{R}$ to every element $R$ of the 3D rotation group.

My question is : why does the position $\hat{\boldsymbol{X}}$ operator transforms like a regular vector when the system is rotated : $$ \langle \hat{X}_i \rangle_{\Psi'} = R_{i,j} \langle \hat{X}_j \rangle_{\Psi} \quad \text{with} \quad |\Psi'\rangle=\hat{R}|\Psi\rangle$$ $R_{i,j}$ being the rotation matrix.