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 $i ).$

Note that $F(H_{1})F(H_{2}) \leq F(X),$ so that $F(H_{1})F(H_{2})$ is, in particular, nilpotent.

Let $q$ be a prime for which $O_{q}(G)$ is non trivial. Then $O_{q}(H_{1}) > O_{q}(G)$ and $O_{q}(H_{1}) \leq X \leq H_{2}.$ Set $U = O_{q}(H_{1}).$ Then $H_{1} = N_{G}(U)$ as $U\not \lhd G$ and $H_{1}$ is maximal.

Now $$O_{q^{\prime}}(H_{1} \cap H_{2}) = O_{q^{\prime}}(N_{H_{2}}(U)) \leq O_{q^{\prime}}(H_{2}).$$

We claim that $O_{q^{\prime}}(H_{1})$ must have order divisible by $p$ if it is non-trivial. For otherwise, $ O_{q^{\prime}}(H_{1}) \leq X \leq H_{1} \cap H_{2}$
and $O_{q^{\prime}}(H_{1}) \leq O_{q^{\prime}}(H_{1} \cap H_{2}) \leq O_{q^{\prime}}(H_{2}),$ so $O_{q^{\prime}}(H_{1}) = O_{q^{\prime}}(H_{2}).$ But then $O_{q^{\prime}}(F(H_{1})) = O_{q^{\prime}}(F(H_{2}))$ and these are both non-trivial, a possibility we have excluded, as it leads to $H_{2} \leq H_{1}.$
This leaves two possibilities to consider. Firstly, it may be that $F(H_{1})$ is a $q$-group for some prime $q$. In that case, $F(G)$ is a $q$-group. We claim that $q<5$ in this case: for otherwise, for $Q \in {\rm Syl}_{q}(X),$ we have $ZJ(Q) \lhd G,$ and
$ZJ(Q)$ is a normal subgroup of $G$ which is invariant under any isomorphism from $H_{1}$ to $H_{2}$.

The other possibility is that whenever $O_{q}(G) \neq 1$, $O_{q^{\prime}}(H_{1})$ has order divisible by $p,$ in which case, $O_{q^{\prime}}(G)$ must have order divisible by $p,$ as $O_{q^{\prime}}(H_{1}) \leq O_{q^{\prime}}(G).$

PROOF THAT THIS CAN'T HAPPEN IN GROUPS Of ODD ORDER: Let $G$ be a finite (solvable) group of odd order. We will prove that if $H_{1}$ is a maximal subgroup of $G$ and $H_{2}$ is a subgroup of $G$ isomorphic to $H_{1},$ then $H_{2}$ is also maximal. By previous work on this problem by myself and Marty Isaacs, we may suppose that $O_{p}(G) = 1,$ where $[G:H_{1}] = p^{a},$ where $p$ is prime and $a$ is a positive integer.
As before, if $\alpha :H_{1} \to H_{2}$ is an isomorphism, we may suppose that nor normal subgroup of $G$ contained in $H_{1}$ is $\alpha$-invariant (note that,if it were, it would be contained in $H_{2}$ also).

Now if $X$ is a finite (solvable) group of odd order, and $O_{p^{\prime}}(X) =1,$ then $ZJ(P) \lhd X,$ where $J$ denotes the Thompson subgroup. This is a Theorem of Glauberman. ZJ(P) is even characteristic, as Glauberman observed. One way to see this is to note that $ZJ(P)$ is the intersection of all Abelian $p$-subgroups of $X$ of maximal order. This follows easily as $J(P^{g}) = J^(P)^{g},$ and so $ZJ(P)$ does not depend on the particular Sylow $p$-subgroup $P$ chosen. It follows, that $ZJ(P)$ centralizes, hence is contained in, each Abelian $p$-subgroup of $X$ of maximal order.
A nilpotent injector $I$ of a finite solvable group $X$ is a maximal nilpotent subgroup of $X$ which contains $F(X).$ A nilpotent injector is unique up to conjugacy in $X,$ and for each prime $q$, the subgroup $O_{q}(I)$ is a Sylow $q$-subgroup of $C_{G}(O_{q^{\prime}}(X)).$
We claim that when $X$ has odd order, and I is a nilpotent injector of $X,$ then $ZJ(I) \lhd X.$ It suffices to prove that $ZJ(O_{q}(I)) \lhd X$ for each prime $q.$
(Here, $J(I)$ denotes the subgroup of $I$ generated by its Abelian subgroups of maximal order).
Let $C = C_{X}(O_{q^{\prime}}(X)).$ Then $O_{q}(I) \in {\rm Syl}_{q}(C)$ and so
$C = O_{q^{\prime}}(C)N_{C}(ZJ(O_{q}(I))$ by Glauberman's theorem. But $O_{q^{\prime}}(C)
\leq O_{q^{\prime}}(X),$ so that $ZJ(O_{q}(I)) \lhd C$. And it is even characteristic, because $O_{q^{\prime}}(C) \times ZJ(O_{q}(I))$ is characteristic. Since $C \lhd X,$ we have $ZJ(O_{q}(I)) \lhd X,$ as required. Hence $ZJ(I) \lhd X.$

Note that $ZJ(I) \leq F(X),$ so that $ZJ(I) \leq T$ for every nilpotent injector $T$ of $X.$ Since $T = I^{x}$ for some $x,$ we see that $ZJ(I) = ZJ(T)$ is independent of the particular nilpotent injector $I$ of $X$ which is chosen.

Now let us turn to our solvable group $G$ of odd order with its maximal subgroup $H_{1}$ and isomorphic subgroup $H_{2}$, both of index a power of $p,$ where $O_{p}(G) = 1.$

We may suppose that $H_{1}$ and $H_{2}$ contain a common Hall $p^{\prime}$-subgroup $M$ of $G,$ possibly after replacing $H_{2}$ by a conjugate. Now $F(H_{1})F(H_{2}) \leq M,$ so that $F(H_{1})F(H_{2}) \leq F(M).$ Let $I$ be a nilpotent injector of $M$ containing $F(M).$ Then $I$ is a nilpotent injector of $H_{1}$ and of $H_{2},$
since it contains $F(H_{1})$ and $F(H_{2})$ and $O_{p}(H_{i}) = 1$ for each $I.$

Also, $I$ is a nilpotent injector of $G,$ since $F(G)$ is a $p^{\prime}$-group.
Thus $ZJ(I) \lhd G.$ How let $\alpha :H_{1} \to H_{2}$ be an isomorphism. Then $T = I\alpha$ is a nilpotent injector of $H_{2},$ and also a nilpotent injector of $G.$
Now $ZJ(I)\alpha = ZJ(I\alpha) = ZJ(T) = ZJ(I).$ This contradicts our assumption that there was no non-trivial $\alpha$-invariant normal subgroup of $G$ contained in $H_{1}.$