I am looking at the following optimization formulation for finding all the eigenvectors of a symmetric matrix $\mathbf{A} \in \Bbb R^{n\times n}$.
$$\min_{\mathbf{\Phi} \in \Bbb R^{n \times n}}\;\operatorname{tr} \left(\mathbf{\Phi}^\top\mathbf{A}\mathbf{\Phi}\right) \quad \text{s.t.} \quad \mathbf{\Phi}^\top\mathbf{\Phi} = \mathbf{I}$$
It appears, for example, in section 3.2 of this tutorial on eigenvalue and generalized eigenvalue problems.
Since $\mathbf{\Phi}$ is a square orthogonal matrix, we have $\mathbf{\Phi}^\top\mathbf{\Phi}=\mathbf{\Phi}\mathbf{\Phi}^\top=\mathbf{I}$. Therefore, the objective value is independent of $\mathbf{\Phi}$:
$$\operatorname{tr}\left(\mathbf{\Phi}^\top\mathbf{A}\mathbf{\Phi}\right) = \operatorname{tr}\left(\mathbf{A}\mathbf{\Phi}\mathbf{\Phi}^\top\right) = \operatorname{tr}\left(\mathbf{A}\right)$$
So how come the solution of this problem is the matrix of eigenvectors? It seems that whatever orthonormal matrix I may use should yield the same minimum value. Moreover, if any orthonormal matrix is a minimizer, it should also satisfy the KKT conditions which result in (according to the link above),
$$\mathbf{A}\mathbf{\Phi}=\mathbf{\Phi}\mathbf{\Lambda}$$
But this is only true for the matrix of eigenvectors, so there seems to be some contradiction here. I understand the problem only exists due to $\mathbf{\Phi}$ being square matrix. Can someone help me resolve it?