SVD of matrix and SVD of its transpose: U and V exchange places

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I have reading the post of SVD of matrix vs. SVD of its transpose for m < n

I am not sure whether I agree with his conclusions:

So, let $\mathbf{A}(m \times n)$ and $\mathbf{B}(m \times n)$ be two real-valued matrices. Minimizing $\| \mathbf{AQ} - \mathbf{B} \|^2_F$ with $\mathbf{Q}$ being a rotation matrix is achieved by

$$\mathbf{Q} = \mathbf{U} \mathbf{V}^T$$

with the SVD of $\mathbf{A}^T\mathbf{B}$ being

$$\mathbf{A}^T\mathbf{B} = \mathbf{UDV}^T \qquad \text{(case 1)}.$$

Generally, I would assume that we get the same results by using the transpose of $\mathbf{A}^T\mathbf{B}$, i.e.

$$\mathbf{B}^T\mathbf{A} = \mathbf{UDV}^T \qquad \text{(case 2)}$$

and calculating

$$\mathbf{Q} = \mathbf{V} \mathbf{U}^T$$

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However, according to my understanding when you take the transpose of a matrix, the $\mathbf{U}$ and $\mathbf{V}$ swap places, then the SVD looks like

$$\mathbf{B}^T\mathbf{A} = \mathbf{VDU}^T \qquad \text{(case 2)}$$

Is this correct?