A problem on a complex matrix complex conjugate to its inverse I came across this problem:

Suppose $B$ is a complex $n\times n$ matrix and $\overline{B}$ the complex conjugate.
  prove  that  $\overline{B}B=1$ if  and  only if  there is  an inverible complex  matrix  $A$  with  $BA=\overline{A}$,  where  $\overline{A}$  is  the  complex  conjugate  of $A$.

Can anyone help me? 
 A: I first want to take the opportunity to emphasise (again, but the answers where I made the comment tend to get deleted) that this is not about unitary matrices, as the operation indicated by a prime is just complex conjugation, no transpose. So for instance
$$
  B=\begin{pmatrix} 1 & \mathbf i \\ 0 & 1 \end{pmatrix}
$$
is a valid choice, for which
$$
  A=\begin{pmatrix} -\mathbf i & c-\mathbf i \\ -2 & 2 \end{pmatrix}
$$
is a valid choice, for any nonzero (to ensure $A$ invertible) real value $c$. To avoid further confusion, I will designate (elementwise) complex conjugation by an overline.
The easy direction is that $\overline BB=I$ whenever for some invertible $A$ one has $BA=\overline A$. Writing $B=\overline AA^{-1}$ one simply gets $\overline BB=\overline{\overline AA^{-1}}\overline AA^{-1}=A(\overline A)^{-1}\overline AA^{-1}=I$.
For the opposite direction I used the suggestion by user1551 (now deleted) that if $\overline B+I$ is invertible, one can take that matrix for$~A$. To make an argument that works for all $B$, note that $\overline B$ can have only finitely many eigenvalues, so one can find a complex number $u$ of norm$~1$ such that $-u^2$ is not an eigenvalue of $\overline B$, so that $\overline B+u^2I$ is invertible, and $A=u^{-1}\overline B+uI$ as well. Then
$$
  BA=B(u^{-1}\overline B+uI) = u^{-1} B\overline B+uB = u^{-1}I+uB
 =\overline{uI+u^{-1}\overline B} = \overline A.
$$
A: I assume that, by the "complex conjugate", you mean elementwise complex conjugate $\overline{B}$. If it was a conjugate transpose $B^*$, then for $A = \left[\begin{smallmatrix}1&1\\2&1\end{smallmatrix}\right]$ we would have $B = A^*A^{-1} = \left[\begin{smallmatrix}3&-1\\1&0\end{smallmatrix}\right]$, which is not unitary, i.e., $B^*B \ne I$.
From $BA = \overline{A}$, we see that $B = \overline{A} A^{-1}$, so
$$\overline{B}B = A \overline{A^{-1}} \overline{A} A^{-1} = A \left( \overline{A} \right)^{-1} \overline{A} A^{-1} = A A^{-1} = I.$$
I was unable to prove the opposite direction. Here is what I got so far.
Let us assume that $\overline{B}B = I$. We write $B$ as $B = C + iD$, where $C = \operatorname{Re}B \in \mathbb{R}^{n \times n}$ is the real part of $B$ and $D = \operatorname{Im}B \in \mathbb{R}^{n \times n}$ is its imaginary part. So,
$$I = \overline{B}B = (C - iD)(C + iD) = C^2 + iCD - iDC +D^2.$$
This means that $CD = DC$, so $C$ and $D$ commute. Also, $C^2 + D^2 = I$.
Edit: Here is an idea:
Let $A = \overline{B^{1/2}}$. Then
$$BA = B \overline{B^{1/2}} = B \overline{B} \overline{B^{-1/2}} = \overline{B^{-1/2}} = B^{1/2} B^{-1/2} \overline{B^{-1/2}} = B^{1/2} (\overline{B^{1/2}} B^{1/2})^{-1} = B^{1/2} = \overline{A}.$$
Obviously, we need to extend the notion of a matrix square root to those matrices that have real negative eigenvalues. I'm in a rush now, so please check this part yourself (I'm not sure that all the steps are correct).
A: Disclaimer: I wrote the following by misreading the question as Complex conjugate transpose rather that only complex conjugate. I keep the answer for this case, it may be complementary. By $A'$, therefore, I mean complex conjugate of $A$.

Proof of the first direction: By your definition, $B$ is a Unitary Matrix. Therefore there is another Unitary matrix $C$ such that $B=CDC'$ where $D$ is diagonal and unitary and $C$ is unitary. Elements of $D$ lie all on unit circle and so there are of the form:
$$
D=\left[
\begin{array} {llll}
e^{iw_1}& 0& \dots& 0\\
0& e^{iw_2}& \ddots & 0\\
0& 0& ...& e^{iw_n}\\
\end{array}
\right]
$$
Define $F$ as follows:
$$
F=\left[
\begin{array} {llll}
e^{-iw_1/2}& 0& \dots& 0\\
0& e^{-iw_2/2}& \ddots & 0\\
0& 0& ...& e^{-iw_n/2}\\
\end{array}
\right]
$$
Note that $DF=F'$ Finally if $A=CFC'$, then we have:
$$
BA=CDC'CFC'=CDFC'=CF'C'=(CFC')'=A'
$$

I really doubt that the information provided here is enough to prove the other side. See that $B(AA'^{-1})=I$ and so $(AA'^{-1})B=I$.
$$
B^{-1}=AA'^{-1},B'=A'^{-1}A \implies B'=A^{-1}B^{-1}A
$$
Hence $B'$ is similar to $B^{-1}$:
$$
B'B=A'^{-1}AA'A^{-1}
$$
If you assume that $A$ is normal, i.e. if $AA'=A'A$ then you have:
$$
B'B=A'^{-1}AA'A^{-1}=(A'^{-1}A')(AA^{-1})=I
$$
Without this assumption you can only prove that $B'$ is similar to $B^{-1}$ and it is called a generalized Unitary matrix which includes all but not only unitary matrices. 
This article about exactly the same question maybe useful:
C.R. DePrima, C.R. Johnson, The range of $A^{−1}A'$ in $GL(n,C)$, Linear Algebra and its Applications, Volume 9, 1974, Pages 209-222
