Find the matrix $B\in \mathbb{C}^{n \times n}$ Suppose $A\in \mathbb{C}^{n \times n}$ be a given unitary matrix then if there exists a non zero matrix $B$ so that $A+B$ is an unitary matrix then $B$ needs to satisfy the equation $A^*B+B^*A+B^*B=0$. How to find the matrix $B$ which satisfy this equation?
 A: Lord Shark the Unknown had it in his comment; if $U \ne A$ is any other unitary,
$U^\dagger U = U U^\dagger = I; \tag 1$
then setting
$B = U - A \ne 0, \tag 2$
we have
$A^\dagger B + B^\dagger A + B^\dagger B = A^\dagger (U - A) + (U - A)^\dagger A + (U - A)^\dagger (U - A)$
$= A^\dagger U - A^\dagger A + U^\dagger A - A^\dagger A + (U^\dagger - A^\dagger)(U - A)$
$=  A^\dagger U - A^\dagger A + U^\dagger A - A^\dagger A + U^\dagger U - U^\dagger A - A^\dagger U + A^\dagger A$
$= A^\dagger U - I +  U^\dagger A - I + I -U^\dagger A - A^\dagger U + I = 0; \tag 3$
so there are indeed very many $B \ne 0$ satisfying
$A^\dagger B + B^\dagger A + B^\dagger B = 0, \tag 4$
one $B \ne 0$ for each unitary $U \ne A$.
In fact, all this stuff works with $U = A$, $B = 0$ as well.
Note Added in Edit, Monday 10 September 2018 12:19 PM PST:  This in response to our OP Saheb's inquiry, expressed below his comment to this answer, as to whether and what might be necessary and sufficient conditions for the existence of $0 \ne B \in \Bbb C^{n \times  n}$ with $A + B$ unitary.  For $n \ge 1$, a sufficient condition is simply the mere existence of unitary $A \in \Bbb C^{n \times n}$; for we may choose $\theta \in (0, 2\pi)$ yielding $e^{i\theta} \ne 1$; then $U = e^{i\theta} A \ne A$ is unitary as well
$(e^{i\theta}A)^\dagger (e^{i\theta}A) = e^{-i\theta}e^{i\theta}A^\dagger A = A^\dagger A = I; \tag 5$
so setting $B = U - A$ we have $A + B$ unitary; $B$ will then satisfy equation (4) as we have seen; and of course, if $B$ satisfies (4) and $A + B$ is unitary, then
$A^\dagger A = A^\dagger A + A^\dagger B + B^\dagger A + B^\dagger B = (A + B)^\dagger (A + B) = I, \tag 6$
so $A$ must be unitary as well.  End of Note. 
