Are the singular values of the transpose equal to those of the original matrix?

Both eigenvalues and singular values are invariant to matrix transpose no matter a matrix is square or rectangular.

The definition of eigenvalues of $A$ (must be square) is the $\lambda$ makes $$\det(\lambda I-A)=0$$ For $A^T$, $\det(\lambda I-A^T)=0$ is equivalent to $\det(\lambda I-A)=0$ since the determinant is invariant to matrix transpose. However, transpose does changes the eigenvectors.

It can also be demonstrated using Singular Value Decomposition. A matrix $A$ no matter square or rectangular can be decomposed as $$A=U\Sigma V^T$$ Its transpose can be decomposed as $A^T=V \Sigma^T U^T$. The transpose changes the singular vectors. But the singular values are persevered.