Let $G=(V,E)$ be a bipartite graph, with partition $V=A \cup B$. Recall that an independent set $I$ of $G$ is a set of vertices sharing no edges.
The independent domination number $i(G)$ is defined to be the minimum cardinality among all maximal independent sets of vertices of $G$.
Suppose $G$ is balanced, i.e. $|A|=|B|$. Is it true that there always exists a balanced independent set $I$ (i.e., $|I \cap A|=|I\cap B|$) of size $|I|=i(G)$?
If not, what additional hypotheses on $G$ would imply such statement?
Suppose that $|A|=|B|=n$, and $G=K_{n,n}$. Then $G$ is balanced, and $i(G)=n$, since the only maximal independent sets are $A$ and $B$, but every independent set in $G$ is a subset of $A$ or of $B$, so $G$ has no non-empty balanced independent set.
I can’t help with the rest, I’m afraid.