No, the following theorem gives counterexamples:
Theorem: $\forall n \ge 3$, $A_{n+1}$ and $S_n$ are normal intermediate subgroups of the inclusion $(A_n \subset S_{n+1})$, but $(A_n \subset S_n \subset S_{n+1} )$ and $(A_n \subset A_{n+1} \subset S_{n+1} )$ are not equivalent.
Lemma: $\forall n \ge 3$, $S_{n}$ is a normal intermediate subgroup of the inclusion $(A_n \subset S_{n+1})$.
Proof: The equality $A_n . g . S_n = S_n . g . A_n$ is clear if $g \in S_n$, but $S_{n+1} = \langle S_n , (n,n+1) \rangle$$S_{n+1} = \langle S_n , \tau \rangle$ with $\tau=(n,n+1)$,
so so it suffices to prove that $A_n . (n,n+1) . S_n = S_n . (n,n+1) . A_n$$A_n . \tau . S_n = S_n . \tau . A_n$, (because $A_{n} \triangleleft S_{n+1}$$A_{n} \triangleleft S_{n}$:
If $\sigma$, $\sigma' \in S_n$ then $a_1\sigma\tau\sigma's_1=\sigma a_2 \tau s_2 \sigma=\sigma s_3 \tau a_3 \sigma = s_4\sigma\tau\sigma' a_4$ ($a_i \in A_n$, $s_i \in S_n$).
Now $S_n= \langle (1,2, \dots , n) , (1,2)\rangle$, and $A_n . (n,n+1) . (1,2) = (1,2) . (n,n+1) . A_n$
for all $n \ge 3$, so it suffices to show that $A_n . (n,n+1) . (1, \dots , n) \subset S_n . (n,n+1) . A_n$.
But $A_n . (n,n+1) . (1, \dots , n) = (n,n+1) . (1, \dots , n) . A_n$,
and $(n,n+1) . (1, \dots , n) = e.(n,n+1) . \sigma_1 = (1,2) . (n,n+1) . \sigma_2$,
with $e, (1,2) \in S_n$ and $\sigma_1 = (1, \dots , n) \in A_n$ if $n$ odd, $\sigma_2 = (1,2). \sigma_1 \in A_n$ if $n$ even.
(Idem, $(1, \dots , n) . (n,n+1) . A_n \subset A_n . (n,n+1) . S_n $)
So, $A_n . (n,n+1) . S_n = S_n . (n,n+1) . A_n $ and then $A_n . g . S_n = S_n . g . A_n$ $\forall g \in S_{n+1}$. $\square$
Proof of the theorem: $A_{n+1}$ is obviously a normal intermediate subgroup because $A_{n+1} \triangleleft S_{n+1}$.
$S_n$ is also a normal intermediate by the previous lemma. Next $(A_n \subset A_{n+1}) \not\sim (S_n \subset S_{n+1})$. $\square$
Remark: For correctness (with the question) note that $A_{n+1} \cap S_n = A_n$ and $A_{n+1}.S_n = S_{n+1}$.