This is a silly question asked just out of curiosity. The question has nothing to do with rotation matrices as far as I know, but I didn't know how else to refer to the following operation.
In linear algebra, suppose I have an $m\times n$ matrix
$$
A = \begin{pmatrix}
a_{11} & \dots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \dots & a_{mn}
\end{pmatrix}
$$
Then suppose I "rotate" the elements of this matrix by 90 degrees to get an $n\times m$ matrix
$$
A^\circlearrowright = \begin{pmatrix}
a_{m1} & \dots & a_{11} \\
\vdots & \ddots & \vdots \\
a_{mn} & \dots & a_{1n}
\end{pmatrix}.
$$
This is similar to the transpose, in that the transpose $A^T$ reflects the matrix $A$ along its diagonal, while $A^\circlearrowright$ rotates it by 90 degrees instead.
The question is just whether this "rotation" operation has any sensible meaning when $A$ is interpreted as a linear transformation. I am guessing the answer is no, but it seemed worth asking anyway, just in case it's something that's been considered for some reason.
A related operation is what we could call the "anti-transpose", which reflects the matrix along the 'other' diagonal and also results in an $n\times m$ matrix:
$$
A^\bot \begin{pmatrix}
a_{mn} & \dots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \dots & a_{11}
\end{pmatrix}
$$
An application of this anti-transpose operation is mentioned in this answer on mathoverflow. (Thanks to Torsten Schoeneberg for pointing that out.)
Best Answer
The rotated matrix can be expressed as $$ A^\circlearrowright = A^TJ, $$ where $J$ is the $m \times m$ matrix with entries $$ J_{ij} = \begin{cases} 1 & i + j = m+1,\\ 0 & \text{otherwise.} \end{cases} $$ We can see that $A^\circlearrowright$ has the same singular values as $A$ by noting that $$ A^\circlearrowright [A^\circlearrowright]^T = A^TA. $$ Interestingly, we also have the relation $$ [A^\circlearrowright]^TA^\circlearrowright = [AA^T]^\perp. $$