If I understand correctly, $\chi(H^\partial)$ is the least number of colors which suffice to color the edges of $H$ so that each vertex is incident with edges of at least two different colors. And it seems to me that $H$ is isomorphic to $(H^\partial)^\partial$ provided that, for any vertices $x,y\in V(H)$, there is an edge $e\in E(H)$ such that $e\cap\{x,y\}=\{x\}$. In that case, I believe the answer is affirmative for all $\kappa,\lambda\ge2$.

If $\kappa\ge4$ and $\lambda=2$, let $H=K_\kappa$ (the ordinary complete *graph* of order $\kappa$).

If $\kappa=2$ and $\lambda\ge4$, let $H=K_\lambda^\partial$.

If $\kappa\ge4$ and $\lambda\ge4$, let $H=K_\kappa\cup K_\lambda^\partial$ (disjoint union).

If $\kappa=2$ and $\lambda=2$, let $H=C_4$.

If $\kappa=3$ and $\lambda=3$, let $H=C_3$.

If $\kappa=3$ and $\lambda=2$, let $H=K_4-e$.

If $\kappa=2$ and $\lambda=3$, let $H=(K_4-e)^\partial$.

If $\kappa\ge4$ and $\lambda=3$, let $H=K_\kappa\cup K_3$.

If $\kappa=3$ and $\lambda\ge4$, let $H=K_3\cup K_\lambda^\partial$.