Consider first the linear system $$ A = X + A X \tag 1$$ where $A$ and $X$ are real symmetric $n\times n$ matrices. Since $A$ and $X$ are real and symmetric, I can take the transpose of both sides to obtain the equivalent system $$ A = X + X A \tag 2$$ whose solution $X$ is the same as in $(1)$. We can also assume both systems are invertible, i.e., $(I-A)^{-1}$ exists. Writing $(1)$ and $(2)$ element-wise as $$ A_{ij} = X_{ij} + \sum_{k=1}^n A_{ik}X_{kj}$$ $$ A_{ij} = X_{ij} + \sum_{k=1}^n X_{ik}A_{kj}$$ I'd like to consider the related system $$ (m_i+m_j)A_{ij} = (m_i+m_j)X_{ij} + m_i \sum_{k=1}^n A_{ik}X_{kj} + m_j \sum_{k=1}^n X_{ik}A_{kj} \tag 3$$ where $m = (m_1 \ldots m_n)^T$ is a vector of positive constants.
Question: Are there solutions to the nonlinear system $(3)$ that are not solutions to the linear systems $(1)$ and $(2)$?
Attempt: By its form, it seems like perhaps I can express $(3)$ as a Sylvester equation. Then I will know it has a unique solution under certain conditions of the eigenvalues of $A$. Since the solution to $(1)$ and/or $(2)$ also solves $(3)$, it must then be the unique solution.
If I define a matrix $M$ by $$ M_{ij} = \frac{m_i}{m_i+m_j}$$ and use the Hadamard product to write $(3)$ as \begin{align} A_{ij} &= X_{ij} + M_{ij} (A X)_{ij} + M_{ji} (X A)_{ij} \\ A &= X + M\circ(A X) + M^T\circ(A X)^T \\ A &= X + M\circ(A X) + (X A)\circ{}M^T \end{align} then I am close to a Sylvester equation, but not quite there.
However, maybe I can learn something from the special case where all $M_{ij}=1/2$ ($m_1 = \ldots = m_n$). In that case, $(3)$ becomes \begin{align} A &= X + \frac{1}{2} A X + \frac{1}{2} X A \\ A &= (I + \frac{1}{2} A) X + X (\frac{1}{2} A) \end{align} which is a Sylvester equation. This gives me reason to think that maybe the general case of $(3)$ can be written in Sylvester form, but the way to do so escapes me.