I believe the answer to be yes, and it follows by some symplectic geometry of Lagrangian intersections.
Let $U$ be the unitary matrix so that $B_j = U A_j$. Without loss of
generality, we will also assume that $A_j = e_j$. This means that
$B_j = U e_j$.
We will identify $\mathbb C^N = \mathbb R^{2N}$.
Then, the first condition on the vector $V$ is that:
$$
|(V, e_j)|^2 = \frac{1}{N}, j=1, \dots, N
$$
This is equivalent to saying that $V = \frac{1}{\sqrt{N}} \sum \mathrm
e^{i \theta_j} e_j$, or, in other words, that $V$ lies in the
Lagrangian torus in $\mathbb R^{2N}$ with the standard symplectic
structure $\sum dx_j \wedge dy_j$, defined by $\{ |x_j|^2 + |y_j|^2 =
\frac{1}{\sqrt{N}} \}$.
The second condition on $V$ is that
$$
|(V, U e_j)|^2 = \frac{1}{N}.
$$
Thus, $U^* V$ also should lie in the torus $L$. Thus, the vector $V$
exists if and only if $L \cap UL$ is non-empty.
(Note the first condition gives automatically that $V$ is a unit vector.)
Right now, I don't see how to take advantage of the linearity in this problem, so I will use an incredibly high powered theory (Floer theory). If I think of a better solution, I will update.
Notice that the action of $U$ on $\mathbb C^N$ induces a map on $\mathbb CP^{N-1}$. Furthermore, if we write $U=\mathrm{e}^{iH}$ for a Hermitian $H$, then $U$ is the time-1 map of the Hamiltonian flow generated by the Hamiltonian $$h(v) = \frac{1}{2} \Re (v, Hv).$$
Finally, we note that $L$ projects to the Clifford torus $L'$ in $\mathbb CP^{N-1}$. It is known for Floer
theoretic reasons (not sure who first proved it... there are now many
proofs in the literature) that the Clifford torus
is not Hamiltonian displaceable, so
there must always exist an intersection point. After normalizing, this lifts to an intersection point in $\mathbb C^N$, as desired.
Note that the Floer homology argument is a very powerful tool. I
suspect that a much simpler proof can be found, since this
doesn't use the linear structure.
EDIT: Apparently my use of the term "Clifford torus" is non-standard. Here is what I mean by it:
Consider $\mathbb CP^{N-1}$ as the quotient of the unit sphere in $\mathbb C^{N}$ by the $S^1$ action by multiplication by a unit complex number (as we have defined here). In the unit sphere there is a real $N$ dimensional torus given by $|z_1| = |z_2| = \dots = |z_N| = \frac{1}{\sqrt{N}}$. The image of this $N$ dimensional torus by the quotient map is an $N-1$ dimensional torus in $\mathbb CP^{N-1}$. Equivalently, it is the torus given in homogeneous coordinates on $\mathbb CP^{N-1}$ by $[e^{i \theta_1}, e^{i \theta_2}, \dots, e^{i \theta_{N-1}}, 1]$.
Hint : Here I have done for $2 \times 2$ matrix.
Let $A = \left(
\begin{array}{cc}
a & 0 \\
0 & b \\
\end{array}
\right)$
be a diagonal matrix with complex entries. Its eigenvalues
are precisely $a$, $b$. Because $A$ is Hermitian, they must be real. Also $A$
is unitary, they must each be of absolute value $1$. There are exactly four
matrices satisfying these conditions:
Let $A_1 = \left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)$, $A_2 = \left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)$, $A_3 = \left(
\begin{array}{cc}
-1 & 0 \\
0 & 1 \\
\end{array}
\right)$, $A_4 = \left(
\begin{array}{cc}
-1 & 0 \\
0 & -1 \\
\end{array}
\right)$
I hope this may help you.
Best Answer
No, a unitary matrix is not its own inverse. It is the inverse of its Hermitian conjugate.
Using the polarization identity, a matrix preserves magnitudes of vectors (all vectors, not just unit vectors) if and only if it preserves the inner product, and that is true if and only if the matrix is unitary:
$$ \langle U x, U y \rangle = \langle x, U^* U y \rangle = \langle x, y \rangle \ \text{for all} \; x,y \; \text{iff}\ \ U^* U = I$$