Given a n-by-n matrix $\mathbf{\phi}$ and a vector $\mathbf{X}$, solve for the two vectors $\mathbf{\Phi}$ and $\mathbf{\Omega}$ that satisfy:
$$ \Phi_i = \sum_{l} \frac{\phi_{il}X_l}{\Omega_l} $$
$$ \Omega_j = \sum_{l} \frac{\phi_{lj}X_l}{\Phi_l} $$
If $\Omega, \Phi$ are a solution then $t\Omega,t^{-1}\Phi$
is also a solution, for any real $t\ne0$, so (this is not a linear
system and) solutions are not unique. Eliminating the $t$ suggests looking
at the products $\Omega_i\Phi_i$.
Take the case where all coordinates of $X$ are nonzero and
the eigenvalues of $\phi$ are known. The following method will find
any solution for which $\Omega_i\Phi_i = \lambda X_i$, if they exist.
I don't know whether other types may exist.
For any vectors $a$ and $b$ I'll use the matlab notation $a.*b$ and $a./b$ for vectors with coordinates $a_ib_i$ and $a_i/b_i$. Your system says $$ \Phi = \phi(X./\Omega), \qquad \Omega = \phi^T(X./\Phi). $$ Denote $u=X./\Omega$ and $v=X./\Phi$, choose an eigenvalue $\lambda$ of $\phi$, and solve for eigenvectors $$ \phi u = \lambda u, \qquad \phi^T v = \lambda v. $$ Then examine $u.*v$. If $u.*v$ is proportional to $X$, rescale $u$ so that $$ \lambda u.*v = X,$$ and you have found a solution $\Omega = X./u, \Phi = X./v$. If not, try another eigenvalue $\lambda$.
