In the following questions we are dealing with subgroups of SYM($\aleph_0$) (the group of permutations of a countable set) with each non identity element having infinite support.
1. Are there two isomorphic subgroups (of SYM($\aleph_0$)) which are not conjugate?
2. Are there subgroups as above, but with transitive natural action on $\aleph_0$?
3. Are there locally finite subgroups as above, with transitive natural action on $\aleph_0$?
Thanks, this is my first time posting here.
For the first question. Let us consider the morphism :
$$f_3:\mathbb{N}\rightarrow \mathbb{N} $$
$$3n\mapsto 3n $$
$$3n+1\mapsto 3n+2$$
$$3n+2\mapsto 3n+1$$
This is clearly a bijection of order $2$ whose support is $(3\mathbb{N})^c$ which is infinite.
On the other hand could also consider :
$$f_2:\mathbb{N}\rightarrow \mathbb{N} $$
$$2n\mapsto 2n+1 $$
$$2n+1\mapsto 2n$$
This is clearly a bijection of order $2$ whose support is $\mathbb{N}$.
Now the subgroup $G$ generated by $f_2$ and $H$ by $f_3$ are both isomorphic to $\mathbb{Z}_2$, nevertheless they cannot be conjugate (otherwise the set of fixed points would be empty for $f_3$ because it is for $f_2$).