Re-edited in view of Derek Holt's comment : Some suggested directions- not a complete answer : Suppose we have $N \lhd G$ with $N \leq H_{1} \cap H_{2},$ and
that there is an isomorphism $\alpha : H_{1} \to H_{2}$ which leaves $N$ invariant.
Then $\alpha$ induces a well-defined isomorphism from $H_{1}/N$ to $H_{2}/N,$ and we may pass to $G/N.$ Hence we assume that there is no non-trivial choice of such an $N$ possible. 

Let $[G:H_{1}] = p^{a}$ for some prime $p$ and integer $a.$   We note the general fact that if $M$ is a subgroup of the finite group $G,$ and $[G:M]$ is a power of $p,$ then $O_{p}(M) \leq O_{p}(G).$ It suffices to prove that $O_{p}(M) \leq P$ whenever $P \in {\rm Syl}_{p}(G).$ But $|PM| = \frac{|P||M|}{|P \cap M|}$, and this is divisible both by  $|G|_{p^{\prime}}$ and by $|P|,$ so that $PM = G$ and $P \cap M \in {\rm Syl}_{p}(M)$. Hence $O_{p}(M) \leq P \cap M,$ as claimed.
   Hence we have $O_{p}(H_{i}) \leq O_{p}(G)$ for each $i.$ Note that $|O_{p}(H_{1})| = |O_{p}(H_{2})|.$ 
   Suppose that $O_{p}(G)  \leq H_{1}.$ Then $O_{p}(H_{2}) \leq O_{p}(G) \leq O_{p}(H_{1}) \leq O_{p}(G) $ and $O_{p}(H_{1}) = O_{p}(H_{2}) = O_{p}(G),$ which is then invariant under any isomorphism from $H_{1}$ to $H_{2}.$ Our assumptions then imply that $O_{p}(G) = 1.$ (Likewise, we can't have $O_{p}(G) \leq H_{2}$ if $O_{p}(G) \neq 1$- the maximality of $H_{1}$ is not essential in the preceding argument).
   Hence, if $O_{p}(G) \neq 1,$ then  $G = O_{p}(G)H_{1}$ and $$|G| = \frac{|O_{p}(G)||H_{1}|}{|O_{p}(H_{1})|} = \frac{|O_{p}(G)||H_{2}|}{|O_{p}(H_{2})|},$$ so $G= O_{p}(G)H_{2}.$ (to be continued)