Matrix equation $XAXBXC=I$ Let $A,B,C$ be unitary matrices. Does there always exist a unitary matrix $X$ such that $$(XA)(XB)(XC)=I,$$ where $I$ is the identity matrix? The quadratic equation $(XA)(XB)=I$ has the solution $A^*(AB^*)^{1/2}$, and I am hoping that the cubic and higher dimensional versions are always solvable.
 A: Here is an argument showing that the answer is 'yes'.  I'll let you check the details and that this result generalizes to all higher degrees.  
Consider the map $f_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f_{ABC}(X) = XAXBXC.
$$
Since the image of this map is compact, if this map were not onto, it would have to have topological degree equal to zero.
Next, since $\mathrm{U}(n)$ is connected, the map $f_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f_{ABC}(X) = XAXBXC
$$
is homotopic to the map $f_{III}$, and hence has the same degree as $f_{III}$. Thus, it suffices to show that the map $f_{III}$ has nonzero topological degree to show that $f_{ABC}$ is surjective.
I claim that the map $f_{III}$ has topological degree $3^n$.  
To prove this, it suffices to compute the local degrees around the pre-images of a regular value.  Let $Y = \mathrm{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})$, where $0<\theta_1<\theta_2<\cdots<\theta_n<\pi$.  Then $Y$ has distinct eigenvalues.  Hence, any solution $X$ to $X^3 = Y$ has distinct eigenvalues and has the same eigendirections as $Y$. Thus, $X$, too, must be diagonal and must be of the form
$$
X = \mathrm{diag}(e^{i\tau_1},\ldots,e^{i\tau_n})
$$
where $3\tau_k \equiv \theta_k\ \mathrm{mod}\ 2\pi$ for $k=1,\ldots, n$.  Thus, there are $3^n$ solutions $X$ to $X^3=Y$.  
I claim that $Y$ is a regular value of the map $f_3:\mathrm{U}(n)\to \mathrm{U}(n)$ defined by $f_3(X)=X^3$ and that $f_3$ is orientation preserving at each solution $X$ of $X^3=Y$.  This follows from the following computation:  Consider the pullback under $f_3$ of the canonical left-invariant form $g^{-1}\mathrm{d} g$.
$$
f_3^*(g^{-1}\mathrm{d}g) = g^{-3}\mathrm{d}(g^3) 
= (I + \mathrm{Ad}(g^{-1})+\mathrm{Ad}(g^{-2}))(g^{-1}\mathrm{d}g)
$$
When one computes the determinant of $\bigl(I + \mathrm{Ad}(X^{-1})+\mathrm{Ad}(X^{-2})\bigr):{\frak{u}}(n)\to {\frak{u}}(n)$ for each solution $X$ of $X^3=Y$, one finds that, because $Y$ has distinct eigenvalues, this determinant is a positive number.  Thus, $Y$ is a regular value of $f_3$, and each of the $3^n$ solutions $X$ to $X^3=Y$ contributes a $+1$ to the topological degree in the usual degree formula.
(By a similar argument, the map $f_k:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by $f(X) = X^k$ has topological degree $k^n$, which is nonzero, so it is necessarily surjective. This answers the higher degree cases as well.)
A: EDIT2: Thanks to Robert Bryant for clearing the unitary fog. My argument on existence is wrong, as it tries to run Brouwer on $U(n)$ (how silly!).
But because the Matlab code that I included seems to often construct a solution, I'm leaving that part of the answer here. The rest, I've excised.

Following Terry Tao's comment, I aim to solve
$$X^{-2}=BXC.$$
Consider now the map
$$\mathcal{G} : U \mapsto B\sqrt{U}^*C.$$
Alas, it is not easy to ensure that this is a fixed-point map (in fact, the $\sqrt$ operation is not even continuous), otherwise we'd be done. However, numerically it seems to yield a unitary matrix $W$ such that upon setting $X^{-1}=\sqrt{W}$ which satisfies $X^{-2}=BXC$.
The following matlab code almost always manages to also construct such a solution! I haven't looked in to see, what particular square-roots does 'sqrtm' in Matlab return for unitary matrices. (analyzing convergence of this algorithm is a question in itself).

function X = unitaryCubeRoot(B,C,maxit)
% Matlab code to solve: X*X*B*X*C=I
  n = size(B,1);
  W = eye(n);
  for k=1:maxit
    W = B*sqrtm(W)'*C;
    X = sqrtm(W)';
    fprintf('%d: %E\n',k, norm(X*X*B*X*C-eye(n)));
  end
end

