Is singular value matrix uniquely determined up to permuting rows and columns.

Given the SVD $$A = U\Sigma V^T$$, is $$\Sigma$$ uniquely determined up to permuting the rows and columns?

My take is that singular value matrix is uniquely determined. Its diagonal elements are square roots of eigenvalues of $$A^TA$$ and $$AA^T$$ (both of which have the same unique set of eigenvalues). Thus, $$A$$ has a unique set of singular values.

Permuting rows and columns of $$\Sigma$$ we just rearrange the singular values along the diagonal of $$\Sigma$$. SVD will continue to hold if we rearrange corresponding vectors of $$U$$ and $$V$$.

However, such reasoning seems to lack rigour.

If an eigenvalue of $$A^T A$$ or $$AA^T$$ has multiplicity greater than 1, the corresponding left and right singular vectors are not unique. In this case, the columns of $$U$$ and $$V$$ can be permuted, but one can also choose entirely different left and right singular vectors.
A simple example. \begin{align*} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}^T \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}^T \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}} \end{bmatrix}. \\ \end{align*}