Are the spectra of $A'A$ and $B'B$ identical if $B$ is a block transpose of $A$? Given block matrices $\mathbf{A}$ and $\mathbf{B}$ which are block transposes of each other, i.e.,
$$\mathbf{A} = \left[ \begin{array}{cccc}
\mathbf{M}_{11} & \mathbf{M}_{21} & \dots & \mathbf{M}_{J1}\\
\mathbf{M}_{12} & \mathbf{M}_{22} & \dots & \mathbf{M}_{J2}\\
\vdots & \vdots & \ddots & \vdots\\
\mathbf{M}_{1K} & \mathbf{M}_{2K} & \dots & \mathbf{M}_{JK}
\end{array} \right]$$
$$\mathbf{B} = \left[\begin{array}{cccc}
\mathbf{M}_{11} & \mathbf{M}_{12} & \dots & \mathbf{M}_{1K}\\
\mathbf{M}_{21} & \mathbf{M}_{22} & \dots & \mathbf{M}_{2K}\\
\vdots & \vdots & \ddots & \vdots\\
\mathbf{M}_{J1} & \mathbf{M}_{J2} & \dots & \mathbf{M}_{JK}
\end{array} \right]$$
where  $\mathbf{M}_{jk} \in \mathbb{C}^{n \times m}$, I'd like show that the non-zero eigenvalues of $\mathbf{A}' \mathbf{A}$ and $\mathbf{B}' \mathbf{B}$ are equivalent, where $A' = conj(A^T)$ is the conjugate transpose.
This is clearly true if $\mathbf{A}=\mathbf{B}'$ (e.g. see here), but can this result be extended to the block-transpose case?
If it matters, $\mathbf{M}_{jk}$ is a non-square lower-triangular toeplitz matrix for my particular problem, though I'm also interested if the result holds in general as well. For what its worth, numerical analysis suggests that this is true, at least for the special lower-triangular toeplitz case.
Shoot - I had computed the spectrum in MATLAB for an initial data set which initially seemed to suggest that this conjecture was true (at least to the limit of numerical noise).  However further investigation of some smaller examples built from random matrices show that this is false...  Nevertheless, the eigenvalue distribution does seem to be remarkably similar, at least for the cases I looked at.  And it is true that trace(A'A) = trace(B'B). If there is anything else interesting that can be said about the relationship between the spectrum of A'A and B'B, I am still very interested.  Thanks!
Counterexample
This is the eigenvalues of the actual dataset I am interested in, computed in MATLAB.
 A: I found a counter-example with no restriction on the blocks, then also with Toeplitz matrices. 
Here is an example that is not a counter-example:
https://www.wolframalpha.com/input/?i=eigenvalues++%7B%7B-2,+3,+5,+7%7D,%7B11,+-2,+13,+5%7D,%7B17,-19,23,29%7D,%7B-31,17,37,23%7D%7D%5ET+%7B%7B-2,+3,+5,+7%7D,%7B11,+-2,+13,+5%7D,%7B17,-19,23,29%7D,%7B-31,17,37,23%7D%7D
https://www.wolframalpha.com/input/?i=eigenvalues+%7B%7B-2,3,17,-19%7D,%7B11,-2,-31,17%7D,%7B5,7,23,29%7D,%7B13,5,37,23%7D%7D%5ET%7B%7B-2,3,17,-19%7D,%7B11,-2,-31,17%7D,%7B5,7,23,29%7D,%7B13,5,37,23%7D%7D
I do not remember the first example that I tried, but in both cases with those Toeplitz matrices, their eigenvalues were really close.
Two things to note, since they have the same trace the first two terms of their characteristic polynomials are the same. Also, a lot of the matrix is the same, just with the block transpose some of the same products will be taken (recall the permutation form of the determinant). I tried to have the terms be prime and large and so little repetition since my earlier example where the eigenvalues were similar had a lot of $1$'s.
It is still kinda striking that the eigenvalues are so close.
A: (Edit: I misread the question. This answer does not apply, but I'll leave it here for reference.)
The answer is "no" in general (it is easy to general random counterexamples), but "yes" if all sub-blocks are $n\times m$ lower triangular Toeplitz.
The eigenvalues of $A^\ast A$ are just the squared singular values of $A$. Each sub-block of $A$ is lower triangular Toeplitz. So, when $m\ne n$, $A$ is guaranteed to have $|m-n|$ zero rows/columns in every $\max(m,n)$ rows/columns (and similarly for $B$). As zero rows/columns do not contribute to nonzero singular values, you may just remove them and assume that each sub-block is square (i.e. $n=m$).
Let $P=P^\top$ denotes the $n\times n$ reversal matrix (i.e. the anti-diagonal matrix obtained by flipping the identity matrix from left to right). Then $M=PM^\top P$. It follows that $B=(P\oplus\cdots\oplus P)A^\top(P\oplus\cdots\oplus P)$. Therefore $B$ is unitarily similar to $A^\top$ and they have identical singular values. But $A$ and $A^\top$ have identical singular values too. Hence the result follows.
