Let $\Phi: R^n \to R^n$ satisfy

$\Phi(x)=u+Ax+Q(x)$, with $x=(x_1, x_2,\ldots, x_n) \in R^n$, $u$ positive vector, $A$ non negative matrix, and $Q(x)$ quadratic mapping with

$Q(x)_i=x_i(k_{i1}x_1+k_{i2}x_2+\ldots+k_{in}x_n)$, where all the $k_{ij}$ are nonnegative and at least one $k_{ij}$ is positive.

Suppose $\Phi(1)=1$, $1$ is the vector each entry being 1

How can I prove that there cannot be two distinct vectors u, v such that u, v are different from the vector 1 and $\Phi(v)=v, \Phi(u)=u$,

$v, u$ are vectors with each entry positive and no greater than 1.

Let $\Phi: R^n \to R^n$ satisfy

$\Phi(x)=u+Ax+Q(x)$, with $x=(x_1, x_2,\ldots, x_n) \in R^n$. $u$ is a given positive vector, $A$ non negative matrix, and $Q(x)$ quadratic mapping with

$Q(x)_i=x_i(k_{i1}x_1+k_{i2}x_2+\ldots+k_{in}x_n)$, where all the $k_{ij}$ are nonnegative and at least one $k_{ij}, 1 \leq i, j \leq n $ is positive.

Suppose $\Phi(\mathbf{1})=\mathbf{1}$, where $\mathbf{1}$ is the vector each entry being 1.

How can I prove that there cannot be two distinct vectors u, v such that u, v are different from the vector $\mathbf{1}$ and $\Phi(v)=v, \Phi(u)=u$,

$v, u$ are vectors with each entry positive and no greater than 1.