Are the singular values of the transpose equal to those of the original matrix? It is well known that eigenvalues for real symmetric matrices are the same for matrices $A$ and its transpose $A^\dagger$. 
This made me wonder:
Can I say the same about the singular values of a rectangular matrix? So basically, are the eigenvalues of $B B^\dagger$ the same as those of $B^\dagger B$?

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.
