I wasn't $100\%$ sure in the accuracy of what I obtained, so may I ask for verification?
Let $A,B\in M_n(\Bbb C)$ be two similar Hermitian matrices. Then $A=P^{-1}BP$.
According to the spectral theorem, every Hermitian matrix is diagonalizable, so $A=U_1^{-1}DU_1$ and $B=U_2^{-1}DU_2$, where $U_1,U_2\in M_n(\Bbb C)$ are unitary and $D=\left(\delta_{ij}\right)\in M_n(\Bbb C)$ is diagonal s. t. $\delta_{ii}\in\sigma(A)=\sigma(B)$.
Since $A$ and $B$ are unitarily similar to $D$, I wanted to write $A$ in the following form: $$A=P^{-1}U_2^{-1}DU_2P$$ Then $U_2P=U_1\implies P=U_2^{-1}U_1$. Since both $U_1$ and $U_2^{-1}$ are unitary, $P$, as a product of two unitary matrices, is also unitary.
Question according to this result:
Are all similar Hermitian matrices unitarily similar?
If this is valid, can it be used in the proof that the matrix representation of a Hermitian operator in an orthonormal basis is a Hermitian matrix? The statement is obvious when, given an arbitrary Hermitian matrix, one is asked to diagonalize it. I thought $P$ could be the transition matrix from one orthonormal basis to another one.
Thank you in advance!
As verifed by @StephenMontgomery-Smith, all similar Hermitian matrices are unitarily similar, indeed.
Howerver, $P$ doesn't necessarily have to be unitary if, e. g., $A=B=0$. Then every invertible $P$ would suffice.
You may see this parallel question I posted on Quora, answered yesterday by Aaron Dunbrack.