(Cross-posted from MSE, with isomorphism replaced by conjugate: https://math.stackexchange.com/questions/4928697/commutativity-of-the-wreath-product?noredirect=1#comment10531931_4928697 )

Let $G$ be a subgroup of the symmetric group $\mathfrak{S}_n$ and $H$ be a subgroup of $\mathfrak{S}_m$. Recall that the wreath product $G \wr H$ is the semi-direct product $G^m \rtimes H$, where $H$ acts on the direct product $G^m$ by permuting components. We can also consider $H \wr G$, the semi-direct product $H^n \rtimes G$. Both can be seen as subgroups of $\mathfrak{S}_{nm}$. For example, $\mathfrak{S}_m \wr \mathfrak{S}_n$ is the stabilizer of the set partition

$$\{ \{ 1, …, m\}, \{m+1, …, 2m\},…, \{(n-1)m + 1, …, nm\}\}.$$

My question is : when are $G \wr H$ and $H \wr G$ conjugated? My suspicion is that it is only the case when $m = n$ and $G$, $H$ are conjugated, or when $G$ and $H$ are both the trivial group. But I can't come up with a proof, nor with a counterexample.

Attempt : First, if $|G|^m|H| \neq |H|^n|G|$, it is obvious, because it is the cardinality of each group. Suppose it is equal, and that there exists an isomorphism $f : G \wr H \to H \wr G$. I suspect that if $f$ is injective for both $G$ and $H$ (considered as subgroups of $G \wr H$), then $|H \wr G|$ must be much bigger than $|G \wr H|$, which is a contradiction. But i don't know if this intuition is good, nor how to formalize it.

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