This is (quite obviously) inspired by this question. Let $C_i$ be symmetric positive definite matrices. Then is it true that there is exactly one symmetric positive definite $X$ such that $F(X) = X^n  \sum_{i=0}^n C_i \circ X^i = 0$, where $\circ$ denotes the Schur (componentwise) product (and exponentiation is with respect to that same product) Notice that unlike in the inspiring question, the Schur product is commutative.
I suspect you mean "is it true that there is exactly one symmetric $X$ such that $F(X) = X^n  \sum_{i=0}^{n1} C_i \circ X^i = 0$", which is more in line with the inspiration question. If you meant something else, I'll just delete this later today. In that case, the answer is no. Take $$ C_2 = \begin{bmatrix} 1 & 2 \\\ 2 & 5 \end{bmatrix} \quad C_1 = \begin{bmatrix} 1 & 1 \\\ 1 & 2 \end{bmatrix} \quad C_0 = \begin{bmatrix} 2 & 2 \\\ 2 & 24 \end{bmatrix} $$ Then we have four polynomials defining the entries in the matrix $X$ (really three by symmetry). The two diagonal entries are unique by Descartes rule of signs, being the real solutions to $z^3 z^2 z  2 =0$, and $z^3 5z^2 2z  24 =0$, so $2$ and $6$  but the off diagonal entries are the solutions to $z^3  2z^2 z +2 = 0$, so they can be $2, 1$ or $1$. Each of these gives a positive definite matrix. 

