Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is

\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}

where the $C_i$ are symmetric positive definite matrices.

If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.

Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?

**EDIT** If the $C_i$ commute with each other, we can simplify matters as follows. Since the $C_i$ commute, they can be simultaneously diagonalized by the same orthogonal matrix $U$; thus, let $C_i=U\Lambda_i U^T$. Suppose for a moment that $\mathcal{G}(X)=0$ does have a solution. Then, pre and post multiplying by $U^T$ and $U$, respectively, we see that this solution must satisfy

\begin{eqnarray*}
U^TX^nU &=& \sum_{i=0}^{n-1} (U^TC_iUU^TX^iU + U^TX^iUU^TC_iU),\\\\
Y^n &=& \sum_{i=0}^{n-1} (\Lambda_iY^i + Y^i\Lambda_i),
\end{eqnarray*}
where $Y=U^TXU$. Now, if *we pick* $Y$ to be diagonal, then we see that indeed, for each diagonal entry we have a separate polynomial that has a unique positive root. Hence, we have a unique diagonal matrix $Y$. But as Mark Sapir alerted me in a comment below, it seems that having a unique diagonal $Y$ (and thus possibly non-diagonal $X=UYU^T$), does not yet rule out the possibility of other solutions.

**Update**
After some hours of struggle *due to the pressure of having posted my question on MO*, under the additional assumption that each term in the sum is positive definite, the answer is "yes". But this assumption seems to be way too strong...