Eigenvalues of matrix product $AB$, where $A$ is a diagonal matrix, and B has a special form? This is a follow-up question to Eigenvalues of matrix product $AB$, where $A$ is a diagonal matrix?.
I have two matrices $A,B$, where $A$ is a diagonal matrix, and $B$ has the form:
$$B_{ij} = f(|j-i|)$$
where $f$ is a generic real-valued function, with $f(x)\ge0$. That is, the value of an entry in $B$ depends only on the distance of this entry from the diagonal.
Suppose I know the eigenvalues of $B$. Is there a way to infer the eigenvalues of $AB$, exploiting the fact that $A$ is diagonal and the special form of $B$?
In the previous question, I did not assume any special properties of $B$. In that general case (as proved in the answer by Robert Israel), there is no clever way to get the eigenvalues of $AB$.
 A: Again, it is not. Consider the following matrices:
\begin{equation}
A=\pmatrix{3&0&0\\0&2&0\\0&0&1}\\
B_1=\pmatrix{1&0&2\sqrt{2}\\0&1&0\\2\sqrt{2}&0&1}\\
B_2=\pmatrix{1&2&0\\2&1&2\\0&2&1}\\
\end{equation}
$B_1$ and $B_2$ have the same eigenvalues $\{1-2\sqrt{2},1,1+2\sqrt{2}\}$, but the eigenvalues of $AB_1$ are exactly $\{-3,2,7\}$ while the eigenvalues of $AB_2$ are approximately
$$\{-3.48,1.51,7.97\}$$

I should note that the abundance of $0$'s in my example above is not particularly important, and if we demanded that $f>0$ (strictly) the answer would not change.
Consider the matrices
\begin{equation}
B_3=\pmatrix{1&1&2\\1&1&1\\2&1&1}\\
B_4=\pmatrix{1&\sqrt{1+\sqrt3}&\sqrt3-1\\\sqrt{1+\sqrt3}&1&\sqrt{1+\sqrt3}\\\sqrt3-1&\sqrt{1+\sqrt3}&1}\\
\end{equation}
Then $B_3$ and $B_4$ have the same eigenvalues $\{-1,2-\sqrt{3},2+\sqrt{3}\}$ but we have that the eigenvalues of $AB_3$ are approximately
$$\{-1.65,0.51,7.14\}$$
while the eigenvalues of $AB_4$ are approximately
$$\{-1.95,0.41,7.54\}$$

How I found these counter-examples
No, I did not pull them out of the hat, haha.
There's a combination of two handy facts at play here; they are well-known and it's a good exercise to try and prove them yourself.

Lemma $\mathbf{(1)}:$ Let $M$ be a square matrix with eigenvalues $\lambda_i, i\in\{1,\dots,n\}$.
  Then
  $$\text{tr}(M^n)=\sum_i\,{\lambda_i}^n$$

${}$

Lemma $\mathbf{(2)}:$ Let $z_1,\dots,z_n\in\mathbb{C}$ and consider the system
  \begin{equation}
\left\{
\begin{array}{cl}
{\lambda_1}+{\lambda_2}+\dots+{\lambda_n}&=z_1\\
{\lambda_1}^2+{\lambda_2}^2+\dots+{\lambda_n}^2&=z_2\\
\dots\\
{\lambda_1}^n+{\lambda_2}^n+\dots+{\lambda_n}^n&=z_n
\end{array}
\right.
\end{equation}
  Up to permutations, there is a unique $(\lambda_1,\dots,\lambda_n)\in\mathbb{C}^n$ which solves the system.

${}$

Corollary: Let $M$ and $N$ be $n\times n$ matrices such that $\text{tr}(M^k)=\text{tr}(N^k)$ for each $k=1,2,\dots,n$.
  Then $M$ and $N$ have the same eigenvalues.

With these in mind, I wrote a general matrix
\begin{equation}
B(a,b,c)=\pmatrix{a&b&c\\b&a&b\\c&b&a}
\end{equation}
and looked for solutions to $\text{tr}\,{B(1,b,c)}^2=\text{tr}\,{B(1,x,y)}^2$ and $\text{tr}\,{B(1,b,c)}^3=\text{tr}\,{B(1,x,y)}^3$.
These are polynomial, low-degree equations on $(x,y)$ in terms of $(b,c)$.
It was just a matter of checking for nonnegative/positive solutions.
