A self-complementary graph $G$ on $n$ vertices is one which is isomorphic to its complementary graph $\bar{G}.$ Some such graphs, for example the cyclic graph on $5$ vertices, have the same degree for each vertex. I'm interested in examples of these. It's fairly easy to show in this situation one needs $n=4k+1.$
If such $n$ are also prime, I have a construction of an example based on a primitive root for $n.$ I'm wondering about composite $n=4k+1.$ Are there any of these for which a constant degree self-complementary graph exists? The first composite case is $n=9,$ next is $n=21,$ etc. Of interest would be an edample in one of these composite cases, or an impossibility proof, even for a few low values of $n.$
I tried it for $n=9$ and couldn't find one. Added note: I would like to assume $G$ is connected. (thisn comes from an example in a comment)
Here is an example of a regular self-complementary graph $G$ of order $9.$
The vertices of $G$ are $u_0,\ u_1,\ u_2,\ u_3,\ v_0,\ v_1,\ v_2,\ v_3,\ w.$
The edges of $G$ are $u_0w,\ u_2w,\ v_1w,\ v_3w,\ u_0u_1,\ u_2u_3,\ v_0v_1,\ v_2v_3,\ u_0u_2,\ v_1v_3,\ u_0v_0,\ u_2v_2,\ u_1v_2,\ u_3v_0,\ u_1v_0,\ u_3v_2,\ u_1v_3,\ u_3v_1.$
You can easily verify that the graph is $4$-regular; the neighborhoods are:
$N(u_0)=\{u_1,\ u_2,\ v_0,\ w\}$
$N(u_1)=\{u_0,\ v_0,\ v_2,\ v_3\}$
$N(u_2)=\{u_0,\ u_3,\ v_2,\ w\}$
$N(u_3)=\{u_2,\ v_0,\ v_1,\ v_2\}$
$N(v_0)=\{u_0,\ u_1,\ u_3,\ v_1\}$
$N(v_1)=\{u_3,\ v_0,\ v_3,\ w\}$
$N(v_2)=\{u_1,\ u_2,\ u_3,\ v_3\}$
$N(v_3)=\{u_1,\ v_1,\ v_2,\ w\}$
$N(w)=\{u_0,\ u_2,\ v_1,\ v_3\}$
The map $u_i\mapsto u_{i+1},\ v_i\mapsto v_{i+1},\ w\mapsto w$ (indices modulo $4$) is an anti-automorphism of $G,$ showing that $G$ is isomorphic to its complement.