Here is the question by @TomH (exact quote from the above):

`Suppose`

$F : [0,1]^n \to [0,1]^n$ `is continuously differentiable and`

$0 < \frac{\partial F_1}{\partial x_i} \leq \dots \leq \frac{\partial F_n}{\partial x_i} < \beta < 1$ `for all`

$i =1,\dots,n$. `Conjecture: there exists unique`

$x^* = F(x^*)$, `and moreover,`

$x_1^* \leq \dots \leq x_n^*$.`

`Proof of the first part is quite straightforward: one can easily verify that`

$F$ `is a contraction mapping and then apply the contraction mapping theorem. I would need some help with the second claim.`

Let me provide a **counter-example** $\ F : [0;1]^2\rightarrow [0;1]^2\ $ in dim 2 (it can be written in $n$ variables too):

Let an auxiliary function $\ g:[0;1]^2\rightarrow \mathbb R\ $ be given as follows:

$$g(x\ y)\,\ :=\,\ \frac14\cdot(x-\frac12)\ +\ \frac12\cdot(y-\frac14)$$

Then $\ -\frac14\le g(x\ y) \le \frac12\ $ for every $\ (x\ y)\in [0;1]^2;\ $ and $\ g(\frac12\ \frac14)\ =\ 0.\ $ Consider $\ F:[0;1]^2\rightarrow [0;1]^2\ $ defined by:

$$F(x\ y)\ :=\ \left(\frac12 + g\left(x\ y\right),\ \frac14 + g\left(x\ y\right)\right)$$

Indeed, the values of the first coordinate of $F$ belong to the interval $\ \left[\frac14;1\right],\ $ and of the second coordinate to $\ \left[0;\frac34\right].\ $ Thus the range of $F$ is in $\ [0;1]^2\ $. Also, the partial derivatives, with respect to $x$, of the two coordinate functions are the same; and the same is true about $y$. Of course both derivatives (constants) belong to a proper closed subinterval of $\ [0;1].\ $ Finally

$$F(\frac12\ \frac14)\ =\ (\frac12\ \frac14)$$

where $\ \frac12 > \frac14.\ $ That's it.