Let $G$ be a finite group then the fundamental theorem of cyclic groups can be formulated as follows:
Theorem: $G$ is cyclic iff it admits no two different subgroups with the same order.
proof: see here p44 together with Lagrange theorem. $\square$
But there is an other characterization of the finite cyclic groups, using the lattice theory:
Theorem: $G$ is cyclic iff its subgroups lattice $\mathcal{L}(G)$ is distributive.
proof: see here theorem 4 p267. $\square$
So we get immediately the following statement for a finite group $G$:
If $G$ admits no two different subgroups with the same order, then its subgroups lattice is distributive.
We ask about a generalization of this statement for an inclusion $(H \subset G)$ of finite groups:
Question: Is it true that if $(H \subset G)$ admits no two different intermediate subgroups $H \subset K \subset G$ with the same order, then its intermediate subgroups lattice $\mathcal{L}(H \subset G)$ is distributive?
Remark: It is checked (by GAP) for $[G:H] \le 31$.
The converse is false because the inclusion $(S_2 \times S_2 \subset S_3 \times S_3)$ is a counter-example (see here).
Best Answer
Say $G=S_n$ and $H$ is a Young subgroup with three orbits, no two of which have the same size and no two of which have sizes summing to $n/2$. Then the only subgroups between $H$ and $G$ should be three Young subgroups with two orbits, no one of which contains any other and no two of which have the same order.