For any $A$ find $B$ such that $x^*B^* A^*BB^*AB x=1$ for all $x\in\{x: \|x\|=1\}$ Proposition:
For any given nonsingular $A\in\mathbb{C}^{n\times n}$, there exists nonsingular $B$, such that
$$x^*B^* A^*BB^*AB x=1\qquad \text{for all}\quad x\in\{x: \|x\|=1\}.$$
This is my own proposition, which I believe probably wrong. However, I want to find a counter-example or prove it.

If $A$ is normal matrix, then we can write $A=U\Sigma U^*$, where $\Sigma$ is diagonal and $U$ is unitary.
Let $B=UB_1$, then
$$x^*B^* A^*BB^*AB x=x^*B_1^*\Sigma^*B_1B_1^*\Sigma B_1x.$$
Using polar decomposition on $B_1$ we get $B_1=PV,$ where $P$ is positive definite and $V$ is unitary. Then
$$x^*B_1^*\Sigma^*B_1B_1^*\Sigma B_1x=y^*P \Sigma^*P^2\Sigma Py,$$
where $y=Vx$ which still satisfies $y\in\{y: \|y\|=1\}$ constraint.
I wanted to choose $P=\Sigma^{-1/2}$, but I don't think it makes $y^*P \Sigma^*P^2\Sigma Py=1$ because of conjugate transpose on $\Sigma$.
 A: First of all, your condition is equivalent to just requiring that $B^* A^*BB^*AB$ is the identity matrix.  Indeed, letting $M=B^* A^*BB^*AB$ then $M$ is a self-adjoint matrix and $(x,y)\mapsto y^*Mx$ is a sesquilinear form.  By polarization, this sesquilinear form is uniquely determined by the function $x\mapsto x^*Mx$, and so it must just be the usual inner product.  It follows that $M$ must be the identity matrix (for instance, by letting $x$ and $y$ range over the standard basis vectors).
So, you are asking for a $B$ such that $B^* A^*BB^*AB=I$, or equivalently a $B$ such that $B^*AB$ is unitary.  Now note that by the spectral theorem, every unitary matrix is diagonalizable by a unitary change of basis, and so by composing $B$ with the latter unitary we can in fact assume that $B^*AB$ is diagonal.  In particular, this means that if $e_1$ and $e_2$ are two distinct standard basis vectors, then the vectors $v=Be_1$ and $w=Be_2$ are linearly independent vectors that satisfy $v^*Aw=w^*Av=0$.
However, for some $A$, no such $v$ and $w$ exist.  For instance, consider $A=\begin{pmatrix}0 & 1 \\ 2 & 0\end{pmatrix}$.  If $v=\begin{pmatrix}a\\b\end{pmatrix}$ and $w=\begin{pmatrix}c\\d\end{pmatrix}$ then $v^*Aw=\bar{a}d+2\bar{b}c$ and $w^*Av=b\bar{c}+2a\bar{d}$.  If both of these are $0$, then conjugating the second and taking appropriate linear combinations gives $\bar{a}d=\bar{b}c=0$.  It is then easy to check that all possible cases would result in $v$ and $w$ being linearly dependent.  So, for this $A$, no $B$ with the property you seek can exist.
