Suppose we have a $t$-uniform hypergraph ($t \ge 3$) $G$, and have $t$ colors available. A question in my research is equivalent to asking what the necessary and sufficient conditions are on $G$ for which no possible vertex coloring of $G$ has every edge rainbow colored. As long as there exists some edge that is not rainbow for every possible coloring of $G$, that's enough.

A sufficient condition I've determined is if $t$ and $G$ contains 3 distinct edges $E_1, E_2, E_3$ for which $|E_1 \cap E_2| \ge t/2$ and $E_3$ contains the symmetric difference of $E_1, E_2$, it is not possible to have all $E_1, E_2, E_3$ be rainbow, and the argument is simple. However, this is not a necessary condition, as other examples of $G$ can be formed. I don't even know of all necessary conditions even when $t = 3$.

Has this problem been studied before?

**Edit**: for $t=3$, an infinite class of $G$'s can be formed: let $k \ge 1$ be any integer, and take $3k+2$ edges $E_1, \cdots, E_{3k+2}$ such that $|E_i \cap E_{i+1}| = 2$ for all $i$ (so a "maximally overlapping path"), and then a final edge $F$ such that $F$ contains the first vertex of $E_1$ and the "last" vertex of $E_{3k+2}$. The choice of $3k+2$ enforces that the same color appears in these two vertices. Then all of $E_1, \cdots, E_{3k+2}, F$ cannot be simultaneously rainbow. Is this the only family of $G$'s for $t=3$?

**Edit 2**: @user36212 rightfully showed that the problem is NP-hard for a number of colors larger than $t$, so I've updated the question for when they are the same.