= Solution
Take the <adjoint matrix> of $M^\dagger\mathbf v_i=\lambda_i\mathbf v_i$ to obtain $\mathbf v_i^\dagger M=\lambda_i^*\mathbf v_i^\dagger$. Evaluating $\mathbf v_i^\dagger M\mathbf u_j$ using this equation and $M\mathbf u_j=\lambda_j\mathbf u_j$ gives
$$
(\lambda_j-\lambda_i^*)\mathbf v_i^\dagger\mathbf u_j=0.
$$
Hence $\boxed{\mathbf v_i^\dagger\mathbf u_j=0\text{ if }\lambda_i^*\ne\lambda_j}$. The proof does not assume that $M$ is a <normal matrix> or <diagonalizable>.
In general the <spectrum> of an <adjoint matrix> is the <complex conjugate> of the original <spectrum>, not necessarily the same set. The equality of the two sets in the question is an additional assumption, possible for a conjugation-invariant <spectrum>; the displayed orthogonality proof only uses the two stated <eigenvalue equations>.
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