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.

Non-obfuscated unprovable programs & many resultant subtleties, Log. Methods Comput. Sci. 12 (2016), no. 2, 2:2, 25 pp. lmcs.episciences.org/1634, which deals with a very similar issue (and relates it to provability). $\endgroup$ – Andrés E. Caicedo Oct 4 '16 at 18:43