Suppose that in $\mathbb{R}^n$, I have an equation $F(x,y,z,\dots,n)=0$ with the property that it is symmetric in the last $(n-1)$ variables $\{y,z,\dots,n\}$, but not in the first variable $x$. For example, if $(1,2,3,\dots,n)$ is a solution, then $(1,3,2,\dots,n)$ is also a solution (found by permuting the second and third numbers), but $(2,1,3,\dots,n)$ might not be.
I now want to solve the following cyclic system of equations:
$$F(x_1,x_2,\dots,x_n)=0$$ $$F(x_2,x_3,\dots,x_n,x_1)=0$$ $$...$$ $$F(x_n,x_1,\dots,x_{n-1})=0$$
Is it true that the solution must satisfy the property that $x_1=x_2=\dots=x_n$?
EDIT: I also want to ensure that all $x_i>0$.
$$(a-4) + (b-4)(c-4) = 0\\ (b-4) + (c-4)(a-4) = 0\\ (c-4) + (a-4)(b-4) = 0 $$
has $ (3, 5, 5)$ as a solution, so no / not necessarily so.