Are self-adjoint / Hermitian operators necessarily orthogonal / unitary?

Question

I feel like self-adjoint / Hermitian operators are the "best" operators, since an operator that is self-adjoint can be orthogonally diagonalized, according to the Spectral Theorem (over the complex number field, an operator only needs to be normal to be orthogonally diagonalizable).

I know that self-adjoint implies normal, but does self-adjoint also imply orthogonal / unitary: $AA^T = A^TA = I$?

Thanks,

Answer 1

Definitely not. A self-adjoint operator just need $A^T=A$. Note the matrix of $T^*$ is the transposed of the matrix of $T$.