Hint:
Given a self complementary graph on $n$ vertices, try to construct a self-complementary graph on $n+4$ vertices.
Elaboration on the hint:
Given a self-complementary graph $G$ with $n$ vertices, add 4 more vertices which already form a 'Z' graph (the self-complementary graph for $n=4$). Connect the vertices of $G$ to some of four newly added vertices. Show that the new graph so formed, is a self-complementary graph on $n+4$ vertices.
Full answer:
Connect every vertex of $G$ to two of the vertices which are the farthest apart on the the n=4 graph (which is basically a path, and you connect all vertices of $G$ to the end points of that path).
This works even when $G$ has only one vertex.
The sum of degrees is $10$, and since the graph is self complementary, by symmetry the degree sequence must be
$$(d_1,d_2, 2,4-d_2,4-d_1) \,,$$
where $d_1, d_2 \in \{ 0,1,2 \}$, and $d_1 \leq d_2$.
It is easy to see that $d_1=0$ is not possible, since then the last degree would be $4$, thus $d_1, d_2 \in \{ 1,2 \}$, and $d_1 \leq d_2$.
Case $1$ $d_1=1, d_2=1$. Your degree sequence is $(1,1,2,3,3)$. Each of the vertices of degree $3$ must be connected to all vertices excluding one end vertex. There is only one graph up to isomorphism.
Case $2$ $d_1=1, d_2=2$. Your degree sequence is $(1,2,2,2,3)$. You have two graphs here: the end vertex is connected to a degree $2$ or $3$.
Case $3$ $d_1=2, d_2=2$. Your graph is connected (why?) and has all degrees $2$. Only 1 possibility.
Best Answer
There is no mistake. We have that $|E(G)| = |E(\overline{G})|$ since $G \cong \overline{G}$ and $|E(G)| + |E(\overline{G})| = \frac{|G|(|G| - 1)}{2}$. Then $|E(G)| = \frac{|G|(|G| - 1)}{4}$.