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 $G= O_{p}(G)H_{2}.$

*Comments on the case $O_{p}(G) = 1$* : In this case, we now know that $O_{p}(H_{i}) = 1$ for each $i.$ Notice that $H_{1}$ and $H_{2}$ each contain a Hall $p^{\prime}$-subgroup of $G,$ and we may suppose that $H_{1}$ and $H_{2}$ have a common Hall $p^{\prime}$-subgroup, say $X$, possibly after replacing $H_{2}$ by a conjugate.
 We again may suppose that no non-trivial normal subgroup of $G$ which is contained in $H_{1} \cap H_{2}$ is invariant under any isomorphism from $H_{1}$ to $H_{2}.$ 
It follows that $O_{p^{\prime}}(H_{i})$ strictly contains $O_{p^{\prime}}(G)$ for each $i$ and that $F(H_{i})$ strictly contains $F(G)$ for each $i$ (in fact, the latter works prime by prime for $O_{q}(H_{i})$ for each $i ).$

Let $M$ be a maximal subgroup of $G$ containing $H_{2}.$ Then $M \neq H_{1}.$ 
Now $O_{p}(M) = 1$ since $X \leq M$ Hence we have $F(M) \leq H_{1}$ and $F(H_{1}) \leq M.$ It may be possible to use Bender methods, though I don't quite see how to do it, as $G$ is not simple. But the subgroup $H_{1}$ has many of the properties of a maximal subgroup of a simple group used in Bender-type proofs.For example,
$H_{1} = N_{G}(O_{q}(H_{1}))$ whenever $O_{q}(H_{1}) \neq 1.$ Similarly, we find that $Z(O_{q}(H_{1}))$ is strictly contained in  $Z(O_{q}(G))$ for all such primes $q$ and 
$H_{1} = N_{G}(Z(O_{q}(H_{1})))$ for all such $q.$ I don't see it as obvious that $M$ need have similar properties, but one might be able to reach the case that $F(H_{1})$ is a $2$-group (or one may have to also allow a $3$-group).