To what extent is the Singular Value Decomposition unique?

In Adam Koranyi's article "Around the finite dimensioal spectral theorem", in Theorem 1 he says that there exist unique orthogonal decompositions.

What is meant here by unique?

We know that the Polar Decomposition and the SVD are equivalent, but the polar decomposition is not unique unless the operator is invertible, therefore the SVD is not unique.

What is the difference between these uniquenesses?

• – glS Jun 12 '19 at 9:05

For distinct singular values, SVD is unique up to permutations of columns of the $$U,V$$ matrices. Usually one asks for the singular values to appear in decreasing order on the main diagonal so that uniqueness is up to permutations of singular vectors with the same singular values.
When singular values are repeated, you have additional freedom of rotating their subspace by an orthogonal matrix $$O$$, e.g. $$U[:,i:j]O$$ and $$V[:,i:j]O$$ for the subset of columns $$i:j$$ which correspond to the same singular value.
The singular values are unique however the columns of $$U$$ and $$V$$ are only unique up to the complex sign (because of how Gram-Schmidt works you're only guaranteed that).