Suppose $A$ is a recursively enumerable subset of $\mathbb{N}$. Let’s call a $\mu$-recursive function $f$ a recursive permutation of $A$, if it is undefined on $\mathbb{N}\setminus A$ and $f|_A$ is a bijection of $A$ onto itself. Let’s denote the set of all recursive permutations of $A$ as $S_R(A)$. It is not hard to see, that $S_R(A)$ forms a group under composition.
Can $S_R(A)$ always be recursively embedded in $S_R(\mathbb{N})$?
If $A$ were decidable, then the question would have been easy: in that case we would have always been able to define the embedding as $f \mapsto g$, where
$$g(x) = \begin{cases} f(x) & \quad x \in A \\ x & \quad x \notin A \end{cases}$$
However, this construction is not possible if $A$ is recursively enumerable, but not decidable. But maybe there is some other way to recursively embed $S_R(A)$ into $S_R(\mathbb{N})$?