I've come across this question while taking a matrix analysis course.
Given: $A,B \in\mathbb{C}^{3 \times 3}$ with $\lambda_1(A) = \lambda_1(B) ,\,\lambda_2(A) = \lambda_2(B),\,\lambda_3(A) = \lambda_3(B)$ (identical eigenvalues) and $\sigma_1(A) = \sigma_1(B) ,\,\sigma_2(A) = \sigma_2(B),\,\sigma_3(A) = \sigma_3(B)$ (identical singular values). are $A,B$ unitary similar? i.e $\exists U:B=U^*AU$ with $U$ being unitary?
I've shown that in the case of $A,B \in\mathbb{C}^{2 \times 2}$, I know that it is true. And I also think this question is related to Specht's theorem.
After taking this problem to Matlab, I've manage to find a counter-example:
$A = \begin{bmatrix} 0 & 1 & 1\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$
$B = \begin{bmatrix} 0 & 1 & 1\\ 0 & 0 & -1\\ 0 & 0 & 0 \end{bmatrix}$
Have the same eigenvalues and singular values. But failed to pass Specht's theorem $tr A^2A^* = tr B^2B^*$ equality.