Let $\mathcal{X} \subset \mathbb{R}^n$ be a finite set and the mapping $\Phi : \mathcal{X} \rightarrow \mathcal{X}$ be defined as follows:
$\Phi(x) := \{y \in \mathcal{X} \mid J(y,x) \leq J(z,x), \, \forall z \in \mathcal{X} \}$
with $J : \mathcal{X} \times \mathcal{X} \rightarrow \mathbb{R}$ two-argument function. I'm not sure about the existence of a fixed point for $\Phi$, but I do not even have a suitable counterexample. Any suggestions/intuitions? Thank you in advance.
Without further assumptions I think you cannot guarantee the existence of a fixed point. Consider, for example, $X= \{1, \ldots, m\}$ for some $m>1$, and let $J(i, j) := |i - (j+1)|$, if $j < m$, $J(i, m) := |i - 1|$. Then $\Phi(j) = \{j+1\}$, if $j < m$, and $\Phi(m) = \{1\}$, and there is no $j \in \{1, \ldots, m\}$ such that $j\in \Phi(j)$.
If you prefer to use a finite subset of $\mathbb{R}^n$ with $n>1$, let $X := \{x_1, \ldots, x_m\} \subset\mathbb{R}^n$ and define the function $J$ as above (with obvious modifications, i.e. $J(x_i, x_j) := |i - (j+1)|$ if $j<m$, etc.).