$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$Let $S\_\infty$ be the group of all permutations of a countable infinite set (enriched with the product Polish topology).
Does there exist a (continuous) group embedding $\phi: S\_\infty\to S\_\infty$ such that $\phi(G)$ is conjugate to $\phi(H)$ for every pair of (closed) isomorphic groups $G, H\leq S\_\infty$?


Note that two subgroups $A, B$ of a group $G$ are conjugate in $G$ if there is $g\in G$ for which $A=gBg^{-1}$.


**Additional note**: It would be stil of interest for me, if there is an embedding $\phi: S\_\infty \to T$ for some Polish group $T$, such that $\phi(G)$ is conjugate to $\phi(H)$ in $T$ for every pair of (closed) isomorphic groups $G, H\leq S\_\infty$?


My first guess was to consider $T$ to be the semidirect product of $S\_\infty$ with $\Aut(S\_\infty)=\Inn(S\_\infty)=S\_\infty$.



Are isomorphic subgroups of the symmetric group conjugate in some larger group?
No, there's no such endomorphism. More generally:



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> **Proposition.** for every infinite set $X$, there exists no set $Y$ and nontrivial continuous homomorphism $f:S\_X\to S\_Y$ for which the following assertion holds: for any two isomorphic discrete infinite cyclic subgroups of $S\_X$, the images $f(S\_X)$ and $f(S\_Y)$ are conjugate.
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Note: every group endomorphism of $S\_\omega$ is continuous (automatic continuity, here essentially due to Dixon-Neumann-Thomas: see Prop 4.3 and Thm 4.4 in [Rosendal's survey](http://homepages.math.uic.edu/%7Erosendal/PapersWebsite/AutomaticContinuity06.pdf)). [The notation $S\_\infty$, which implicitly assumes that every infinite set is countable (Cantor disproved this) is inconvenient.]


Let $c$ be any nontrivial permutation of $X$ consisting only of infinite cycles (at least one) and fixed points. Let $c'$ be any permutation of $X$ with both at least one infinite cycle and finite cycles of all sizes.


Since both $c$ and $c'$ have an infinite cycle, the cyclic subgroups $\langle c\rangle$ and $\langle c'\rangle$ are infinite and discrete (hence closed).



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> **Proposition.** For every set $Y$ and homomorphism $f:S\_X\to S\_Y$, the subgroups $\langle f(c)\rangle$ and $\langle f(c')\rangle$ are not conjugate.
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> More precisely, $f(c)$ has no finite cycle of size $\ge 2$, while (unless $f$ is the trivial homomorphism) $f(c')$ has finite cycles of unbounded size.
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(This immediately implies the first proposition, in a strong sense since the embedding property fails for a given pair of isomorphic closed subgroups.)


**Lemma** Let $H$ be an open subgroup of $S\_X$. Then there exists a finite subset $F$ of $X$ such that $H$ is trapped between the pointwise and global stabilizer of $F$. $\Box$


[This is classical. The main case is when $H$ is transitive, in which case the conclusion is $H=S\_X$. The general case easily follows.]


Through $f$, we can view $Y$ as an $S\_X$-set. For both statements it enough to assume that $Y$ is a transitive $S\_X$-set (through $f$). Hence $Y=S\_X/H$ for some open subgroup $H$.


For the statement about $c$ (no nonsingleton finite cycle), we can use the lemma and find an open finite index subgroup $H'$ of $H$, such that $H'$ is the pointwise stabilizer of a finite subset. We can identify $S\_X/H'$ with the set of pairwise distinct $n$-tuples in $X^n$ for the product action, for some $n\ge 0$. Indeed we see that $c$ has no nonsingleton finite cycle on $\omega^n$. Since the equivariant surjection $S\_X/H'\to S\_X/H$ is finite-to-one, we deduce that $c$ has also no nonsingleton finite cycle on $S\_X/H$.


For the statement about $c'$ (existence of unbounded finite cycles), we can use the lemma again, now defining $H'$ as finite index overgroup of $H$, such that $H'$ is the global stabilizer of a finite subset $F$. If $F$ is empty, we have $H=H'=S\_X$ and the statement is void. Assume now $F$ nonempty. We can identify $S\_X/H'$ with the set of $n$-elements subsets of $X$ for some $n\ge 1$. Since $c'$ admits a $kn$-cycle for every $k\ge 2$, we see that $c'$ has a $k$-cycle for its action on $S\_X/H'$. Hence $c'$ has a $kk'$-cycle for some $k'\ge 1$ on $S\_X/H$.


Using automatic continuity, let me mention, out of curiosity:



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> **Corollary.** For $X=Y=\omega$ and $c,c'$ as above, the HNN extension of $S\_\omega$ over the isomorphism $\langle c\rangle \to \langle c'\rangle$ mapping $c$ to $c'$, has no faithful action on any countable set.
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(Indeed, finding groups of at most continuum cardinal with no faithful action on any countable set has been an open problem long ago, and while various examples are now known, such a construction seems to be a new one. Note that no group topology is involved in the above statement.)




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This argument doesn't address general Polish groups as targets. However it suggests that it already makes sense to attempt to make discrete subgroups conjugate, even infinite cyclic ones.


