Solving a set of non-linear matrix equations Consider the following set of equations
$$\begin{cases}PAQ^{-1}&=T \\
QBR^{-1}&=T\\
RCP^{-1}&=T,
\end{cases} $$
where A,B,C and T are known real-valued $3\times3$ matrices and P, Q, R are the unknown (real valued, $3\times3$ sized, invertible matrices). 
Is it there an analytic solution for this set of equations? 
If not, then consider one or more of the following relaxations:


*

*$T = I_3$ or $T=\begin{bmatrix}1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$

*A, B, C are invertible

*Numeric solutions are also acceptable

 A: We can multiply the given equations in different orders to get: 
$T^3 = PABCP^{-1} = QBCAQ^{-1} = RCABR^{-1}$
Thus, in order for a solution to exist, we need $ABC$, $BCA$, $CAB$, and $T^3$ to be similar matrices. 
If that is indeed the case, we have the following: 
$P(ABC) = T^3P$
$Q(BCA) = T^3Q$
$R(CAB) = T^3R$
Using the vectorization operation, these equations become:
$((ABC)^T \otimes I - I \otimes T^3)\text{vec}(P) = 0$
$((BCA)^T \otimes I - I \otimes T^3)\text{vec}(Q) = 0$
$((CAB)^T \otimes I - I \otimes T^3)\text{vec}(R) = 0$
where $\otimes$ denotes the Kronecker product. 
This gives you a $9 \times 9$ linear system for each matrix $P,Q,R$. Each system should be rank deficient, so you should get at least a one dimensional nullspace of solutions for each matrix. (Which makes sense because if $(P,Q,R)$ is a solution, so is $(kP,kQ,kR)$ for any non-zero constant $k$). 
After solving one of these equations for one of $P$ or $Q$ or $R$, the other two are uniquely determined using the original three equations. The question that remains is can we get a two (or larger) dimensional nullspace for all of these equations, and what to do if that happens. 
