A conjecturally easier version of matrix Descartes rule of signs.

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 (component-wise) 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 $F(X) = X^n - \sum_{i=0}^{n-1} C_i \circ X^i = 0$", 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.