Let $A,B\in M_n$. Show that $|x^\ast Ay|=|x^\ast By|$ for all $x,y\in \mathbb{C}^n$ iff $A=e^{i\theta}B$ for some $\theta\in \mathbb{R}$.
The one side is easy to prove by plugging $e^{i\theta}B$ in and taking out $e^{i\theta}$, and $|e^{i\theta}|=1$. For the other side, I know that I may use Toeplitz decomposition to write $|x^\ast Ay|=|\frac{1}{2}x^\ast (A+A^\ast)y+i\frac{1}{2i}x^\ast(A-A^\ast)y|$, I might need to check $A+A^\ast=\sin\theta B$ and $\frac{1}{2i}(A-A^\ast)=B$, but I'm stuck here. Plus, I might also write $|x^\ast Ay|$ to $|\sum_{i,j=0}^na_{i,j}\bar{x_i}y_j|$ and $|x^\ast By|$ to |$\sum_{i,j=0}^nb_{i,j}\bar{x_i}y_j|$, but I have no ideas how to prove in this way either.
Strategy talk: the condition is that the norms must be equal for all $x, y$. This is extremely strong. So, it may be a good idea to proceed by contraposition, presuming the elements of the matrices don't satisfy our target statement somewhere, and try to construct a pathological case that demonstrates the antecedent's failure. In fact, while trying to do such a proof, you might find one that reads easier directly, with your pathological $x,y$ supposed from the beginning; that's what happened to me here.
If $|x^*Ay| = |x^*By|$ in general, then it must also hold for the particular vectors $x = a_{ij}\hat{e}_i$ and $y = a_{ij}\hat{e}_j$ for each $0 \leq i,j \leq n$ (chosen to "pick out" the ${ij}$ term of the matrix). This yields $x^*Ay = a_{ij}^*a_{ij}^2 = |a_{ij}|a_{ij}$. The norm of this is, using multiplicativity, $|a_{ij}|^2$. One also has $x^*By = a_{ij}^*b_{ij}a_{ij} = |a_{ij}|b_{ij}$, whose norm is $|a_{ij}||b_{ij}|$; equating these, $$|a_{ij}|^2 = |a_{ij}||b_{ij}| \Leftrightarrow |a_{ij}| = |b_{ij}|$$ (In the case $a_{ij} = 0$ it can be shown $b_{ij} = 0$ by taking $x = \hat{e}_i$ and $y = \hat{e}_j$)
Complex numbers whose norm is identical only differ by a phase. Can you see how to use this kind of argument to show that it must be the same phase for any two matrix elements?
P.S. My initial attempts to finish this were frustrated, but I've got other things to attend to and I don't want to give the whole answer in case this is a homework problem; try working with a linear combination, since you'll need interactions between two terms. The triangle inequalities may also be good to keep in mind.