Computability of sets of fixed point values Recall that Rogers' fixed point theorem states that, if $f$ is a total computable function, then it has a fixed point value, i.e., there is an index $i$ such that $\varphi_i\simeq\varphi_{f(i)}$.
My question is what is known about the recursiveness and recursive enumerability of the set of fixed point values of $f$.
I can prove, for instance, that if the orbit (under $f$) of each index has finite complement (with respect to $\mathbb N$), then the set of fixed point values of $f$ is not recursive. The techniques I used have been around since the works of Kleene, hence my results should be well known. Still, I was unable to find any discussion on this topic in the relevant literature, except for a couple of exercises on Rogers' and Odifreddi's monographs, which seem to hint at a well-developed theory, but provide no specific references.
Any pointers?
EDIT:  to make my question more concrete, let us focus on the function $f=\lambda x.x+1$. I know that the set $F$ of the fixed point values of $f$ is not recursive, for any admissible numbering of computable functions. What about recursive enumerability? Are there two admissible numberings that make $F$ r.e. and not r.e., respectively?
 A: In general, the question of whether a given program $i$ is a
fixed-point with respect to a given computuble function $f$, has
complexity $\Pi^0_2$. And for some functions, it is
$\Pi^0_2$-complete.
First, it is easy to see that the assertion that $i$ is a fixed
point with respect to $f$, meaning that $\varphi_i=\varphi_{f(i)}$, has
complexity at most $\Pi^0_2$, since $i$ is a fixed point just in
case for every converging instance of one of the functions, there
is a corresponding converging instance of the other with the same
output value.
Conversely, let me provide a computable function $f$ for which the
fixed-point set is $\Pi^0_2$-complete. Fix a $\Pi^0_2$-universal
set $U$, where $x\in U$ if and only if $\forall k\
\exists j\ A(x,k,j)$, where $A$ is $\Delta_0$. Define $f$ as
follows. For each $i$, let $f(i)$ be a program undertaking the
following procedure: first, let $x=\varphi_i(0)$; now, on input
$k$, search for $j$ for which $A(x,k,j)$, and output $x$ if found;
otherwise keep searching.
For any $x$, let $e_x$ be a program that is known to compute
constant value $x$. Observe that $x\in U$ just in case $f(e_x)$
computes the constant value $x$. Thus, $x\in U$ if and only if
$e_x$ computes the same function as $f(e_x)$, which is to say, if
and only if $e_x$ is a fixed point with respect to $f$.
So for this function, the fixed-point set is $\Pi^0_2$-complete. In
particular, it is not c.e.
Update. Meanwhile, let us consider your updated question,
focussed on the particular function $s(x)=x+1$. I claim that there
is an admissible enumeration of computable functions for which the fixed-point question for this function is
is not c.e. Suppose for example that we have an encoding where odd
numbers always encode the empty function. (Many of the naturally
occurring encodings of Turing machines or whatever have this
property, if you use, say, prime powers for sequence encoding or
$2^n(2m+1)$-type pairing functions, since odd numbers wouldn't code
things as usual, and the corresponding enumerated computable
function $\varphi_k$ for $k$ odd would be the empty function by
default.) Thus, a fixed point $\varphi_e=\varphi_{e+1}$ for the
successor function in this case would involve at least one odd
index, and so it would be the empty function. Thus, an index $e$ is
a fixed point if and only if $e$ is even and $\varphi_e$ is the
empty function, or $e$ is odd and $\varphi_{e+1}$ is the empty
function. This is easily seen to be co-c.e. and $\Pi^0_1$-complete,
using methods as in my answer above. Basically, it is equivalent to
the emptiness problem, which is $\Pi^0_1$-complete. In particular, it is not c.e.
