Fixed point for a map from $\{0,1\}^N$ to itself

Question

Let $N\geq2.$ Let $F$ be a function from $\left\{ 0,1\right\} ^{N}$ to itself dreceasing for the product order defined by $$ (x_1,x_2,\ldots,x_N)\leq (y_1,\ldots,y_N)\ \text{ if and only if for all }i,\ x_i\leq y_i $$

Here, $F$ being decreasing means $$x\leq y \Rightarrow F(y)\leq F(x)$$

Suppose moreover that the $i^{th}$ component of $F$ does not depend on the $i^{th}$ variable.

Is it true that $F$ has a unique fixed point ?

Answer 1

For $N = 3$ there are exactly $58$ counterexamples.

$54$ of them have two fixed points. E.g.: \begin{eqnarray} 000 &\mapsto& 110\\ 100 &\mapsto& 100\\ 010 &\mapsto& 010\\ 110 &\mapsto& 000\\ 001 &\mapsto& 000\\ 101 &\mapsto& 000\\ 011 &\mapsto& 000\\ 111 &\mapsto& 000\\ \end{eqnarray}

$2$ of them have three fixed points. E.g.:

\begin{eqnarray} 000 &\mapsto& 111\\ 100 &\mapsto& 100\\ 010 &\mapsto& 010\\ 110 &\mapsto& 000\\ 001 &\mapsto& 001\\ 101 &\mapsto& 000\\ 011 &\mapsto& 000\\ 111 &\mapsto& 000\\ \end{eqnarray}

2 of them have zero fixed point. E.g.: \begin{eqnarray} 000 &\mapsto& 111\\ 100 &\mapsto& 101\\ 010 &\mapsto& 110\\ 110 &\mapsto& 100\\ 001 &\mapsto& 011\\ 101 &\mapsto& 001\\ 011 &\mapsto& 010\\ 111 &\mapsto& 000\\ \end{eqnarray}

Answer 2

No. Consider $$F(0)=7,\ F(1)=5,\ F(2)=3,\ F(3)=1$$ $$F(4)=6,\ F(5)=4,\ F(6)=2,\ F(7)=0$$ where a number represents its base-2 expansion, e.g. 6 represents $(1,1,0)$.