Quoting an answer I posted some time ago on MathSE:
Take a complete graph with vertex set $V$ and edge set $E={V\choose2}$. Let $\alpha$ be any permutation of $V$ in which the length of each cycle is a multiple of $4$, except for at most one $1$-cycle. Of course, such permutations exist if and only if $|V|\equiv 0$ or $1\pmod 4$. (N.B. The word "cycle" is used here in its group theory sense, not its graph theory sense!)
Let $\beta$ be the permutation of $E$ induced by $\alpha$. Observe that $\beta$ contains only cycles of even length. Color the edges in each cycle alternately black and white. The graph consisting of the black edges is self-complementary; the permutation $\alpha$ is an isomorphism between the black graph and the white graph.
Example. To construct self-complementary graphs of order $5$, take $V=\{a,b,c,d,e\}$ and let $\alpha=(a\;b\;c\;d)(e)$ so that $\beta=(ab\;bc\;cd\;ad)(ac\;bd)(ae\;be\;ce\;de)\;$. If we choose the edges $ab,cd$ and $ac$ (i.e. "color them black") we get a $4$-point path $P_4$. Now we can choose the edges $be,de$ obtaining the self-complementary graph $C_5$, or else we can choose $ae,ce$ obtaining the other self-complementary graph of order $5$, the bull graph.
An alternative proof is often given in graph theory textbooks, e.g., quoting Exercise 1.1.31 on p. 17 of Douglas B. West's Introduction to Graph Theory, second edition:
Prove that a self-complementary graph with $n$ vertices exists if and only if $n$ or $n-1$ is divisible by $4.$ (Hint: When $n$ is divisible by $4,$ generalize the structure of $P_4$ by splitting the vertices into four groups. For $n\equiv1$ mod $4,$ add one vertex to the graph constructed for $n-1.$)