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Let $\Pi$ be the group of recursive permutations (i.e. bijections ${\bf N} \to {\bf N}$ that are computable. Let $\phi_e : e \in {\bf N}$ be an enumeration of the recursive partial functions.

I think the set $\{e \in {\bf N} : \phi_e \in \Pi \}$ is not recursively enumerable. It is easy to see that it is not recursive by reducing it to the halting problem (on input n simulating running $\phi_n(n)$ then return $n$ when it halts).

Furthermore, I wish to show that there is no recursively enumerable subset such that $\{ \phi_e \mid e \in P' \}$ is equal to $\Pi$.

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For the first part, suppose $f$ enumerates $\Pi$. By the Recursion theorem, we can build a partial recursive function $\varphi_e$ whose code "refers to" $f$'s behavior on input $e$. Now, the only constraint on $f$ is that $f(k)$ halts iff $\varphi_k$ is a permutation; so, our $\varphi_e$ should "look like" a permutation until (if ever) it sees $f(e)$ halt, and then "stop being a permutation."

HINT: the simplest permutation is the identity . . .

For a further exercise, modify this idea to show that $\Pi$ is, in fact, not even $\Sigma^0_2$! It is, in fact, $\Pi^0_2$ complete.

For the second part, first think about the classical proof that there are uncountably many permutations of $\mathbb{N}$: given a family $\pi_i$ of permutations, do you see how to diagonalize against them to produce a new permutation $\rho$? Now, that argument effectivizes in a natural way . . .

Note that this provides another argument that $\Pi$ is not recursively enumerable. But, the argument in the first part is of a kind that it is very good to know.

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    Let X = even integers, we set $\rho(2n) in X$ different from $\rho(2m)$ for $m < n$. Then set $\rho(2n-1)$ to be the $n$th element of $N - \{\pi(2n) : n \in{\bf N} \}$. Doesn't effectivise well since one would need to calculate $\pi(2n)$ for all evens first. – 2017-02-16
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    @jay_hathaway You're doing too much work. Here's a hint: let's say I want to make a permutation $\rho$ which is different from a permutation $\pi$. Well, either $\pi(1)=2$ or not; so I can make $\rho$ different from $\pi$ by setting $\rho(1)=2, \rho(2)=1$ if $\pi(1)\not=2$, and $\rho(1)=1, \rho(2)=2$ if $\pi(1)=2$. Note moreover that this "confines" my action to $1$ and $2$ - I've declared nothing about e.g. $\rho(17)$. Do you see how to handle *many* permutations at once this way? And how to effectivize it? – 2017-02-16