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Let $S_n$ be the symmetric group. let $H=S_{d_1}\times\cdots \times S_{d_k}$ such that $d_1+\cdots + d_k=n$.

What does the following statement means:

every two embeddings of $H$ in $S_n$ are conjugate.

In my opinion this means that changing the order of $d_i$ gives conjugate subgroups. For example $S_{d_1}\times S_{d_2}\times S_{d_3}$ is conjugate to $S_{d_2}\times S_{d_1}\times S_{d_3}$ and to $S_{d_3}\times S_{d_1}\times S_{d_2}$ and so on ...

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    In general, when faced with a term you don't know, it's best to ask the source (whether text or person) for a definition rather than assuming a meaning arbitrarily.2011-07-03

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I think that your interpretation is correct. It is certainly not true that any subgroups of $S_n$ that are abstractly isomorphic to $H$ are conjugate. The most famous example is that of two non-conjugate $S_5$ sitting inside $S_6$. So the statement must refer to the canonical embedding of $H$ in $S_n$, in which case the only way I can think of to obtain different embeddings is by permuting the $d_i$ (and then the statement is almost trivially true).

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    @palio See e.g. JA Todd, The Odd Number 6, Math. Proc. Camb. Phil. Soc. 41 (1945) or http://en.wikipedia.org/wiki/Automorphisms_of_the_symmetric_and_alternating_groups#Exotic_map_S5_.E2.86.92_S62011-07-03