TITLE: Fixed point for a map from $\{0,1\}^N$ to itself
QUESTION [7 upvotes]: 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 ?

REPLY [6 votes]: 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}