Let $(V,\|\cdot\|,\leq)$ be a complete, normed and partially ordered vector space. Moreover, let $F:V\rightarrow V$ be a contraction, i.e. there exists $M\in[0,1)$ with
$$\|\;F(x) - F(y)\;\| \;\leq\; M\;\|\;x-y\;\|$$
for all $x,y\in V$.
Intuitively , the following should hold:
For $m\in V$ with $m\geq 0$, define the translated function $$F_{m}:\; V\;\rightarrow\; V,\; x\;\mapsto\; F(x)+m$$ (which is also a contraction) and denote by $\bar{x}_{m}\in V$ the unique fixed point of $F_{m}$. Then, $$\bar{x}_{0} \;\leq\; \bar{x}_{m} \quad.$$
Unfortunately, I am missing a starting point for a rigorous proof. Maybe somebody has a good idea to get me started. Thanks in advance!
EDIT: Noah's answer shows that this is not true in general. But maybe somebody has an idea to proof the conjecture for the case $V=\mathbb{R}^{n}$, $\|\cdot\|$ the euclidean norm, and $x\leq y$ iff $x_{i} \leq y_{i}$ for $i=1,\ldots,n$ (componentwise order).
This is definitely not true in general. Take $V=\mathbb{R} $ with the usual norm and partial ordering $x\le y$ iff $x=0$ or $x=y$. This is clearly reflexive and antisymmetric, and is transitive after a little casework. Now, $f(x)= \frac{1}{2}x +\frac{1}{4} $ is a contraction map with fixed point $\frac{1}{2} $, but $f_{\frac{-1}{4}} $ has fixed point $0$, which under our partial ordering is not greater than or equal to $\frac{1}{2} $.