I am having some trouble seeing why a graph of 4 nodes and 2 edges is not self-complementary such that $G$ is isomorphic to $\overline{G}$ (G complement) (please see the attachment below). I know that the number of edges in a self-complementary graph must be $\frac{1}{4}n(n-1)$, (half of that of a $K_n$ graph) and so for a graph of 4 nodes, the number of edges must be 3. However, why is the graph attached not self-complementary?
The number of edges, nodes, the degree sequence, length of cycles, etc., are the same, and could I not map $G$ and $G$ complement like: $f(d) = b$; $f(b) = d$; $f(a) = a$; $f(c) = c$?
I must be missing something about isomorphism here - if someone could point it out, that'd be great - thanks so much!
As David pointed out in a comment, the second graph is simply not the complement of the first.