$C$: real symmetric matrix ($C \in \mathbb{R}^{n \times n})$
$C=PDP^{-1}$ (orthogonal diagonalization of $C$)
$P$: orthogonal matrix whose column vectors are eigenvectors of $C$
$D$: diagonal matrix whose main diagonals are eigenvalues of $C$
Is orthogonal diagonalization $C=PDP^{-1}$ is unique for each $D$(each eigenvalues' permutation)?
(Add)
Counterexample:
$
\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$
$
\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
$
Eigenvalues and eigenvectors of
$
\begin{pmatrix}
a & 0 \\
0 & b
\end{pmatrix}
$
are
$
\lambda_1=a, x_1=(t, 0) \\
\lambda_2=b, x_2=(0, k) \\
t, k \in \mathbb{R}
$
So orthogonal diagonalization for each eigenvalues' permutaion is not unique.