Satisfying the new Condition 1, If $|X|\geqslant4$, let ${\cal F}$ consist of all subsets of $X$ with two elements.

Condition 1 can be weakened. For example, if $X=\{a_{i,j}:1\leqslant i,j\leqslant 3\}$, then the following 12 subsets, each of size 3, satisfy the conditions.
$$
\{a_{1,1},a_{1,2},a_{1,3}\},\{a_{2,1},a_{2,2},a_{2,3}\},\{a_{3,1},a_{3,2},a_{3,3}\}
$$
$$\{a_{1,1},a_{2,1},a_{3,1}\},\{a_{1,2},a_{2,2},a_{3,2}\},\{a_{1,3},a_{2,3},a_{3,3}\}
$$
$$
\{a_{1,1},a_{2,2},a_{3,3}\},\{a_{1,3},a_{2,2},a_{3,1}\}
$$
$$
\{a_{1,1},a_{2,3},a_{3,2}\},\{a_{1,3},a_{2,1},a_{3,2}\},\{a_{1,2},a_{2,1},a_{3,3}\},\{a_{1,2},a_{2,3},a_{3,1}\}
$$

It seems likely that we can permit any minimum size for sets in ${\cal F}$.