Since $$\alpha(\overline{K_n \boxtimes \overline{K_n}}) = n$$ and $$\alpha(K_n \boxtimes \overline{K_n}) = n$$ it follows that the family of complete graphs satisfies your assumptions.

Since $$\alpha(\overline{K_n \boxtimes \overline{K_n}}) = n$$ and $$\alpha(K_n \boxtimes \overline{K_n}) = n$$ it follows that the family of complete graphs satisfies your assumptions.

To see the first equality observe that $K_n \boxtimes \overline{K_n}$ is the disjoint union of $n$ copies of $K_n.$ Hence in the complement every disjoint $K_n$ is an independent set any pair of vertices from disjoint copies of $K_n$ is adjacent.