For any $n\in\mathbb N,$ here's how you can construct a self-complementary graph of diameter $2$ and order $4n+1.$
Choose a graph $G$ of order $n.$
Start with a $C_5$ graph, with vertices $A,B,C,D,E$ and edges $AB,BC,CD,DE,EA.$
Replace each of the vertices $B$ and $E$ with a copy of $G,$ and replace each of the vertices $C$ and $D$ with a copy of the complementary graph $\overline G.$
More precisely: The graph has vertex set $A\cup B\cup C\cup D\cup E$ where $A,B,C,D,E$ are disjoint sets and $|A|=1$ and $|B|=|C|=|D|=|E|=n.$ The induced subgraphs on $B$ and $E$ are isomorphic to $G,$ the induced subgraphs on $C$ and $D$ are isomorphic to $\overline G.$ There are edges joining all vertices in $A$ to all vertices in $B,$ all vertices in $B$ to all vertices in $C,$ all vertices in $C$ to all vertices in $D,$ all vertices in $D$ to all vertices in $E,$ and all vertices in $E$ to all vertices in $A.$ On the other hand, there are no edges between $A$ and $C,$ or between $C$ and $E,$ or between $E$ and $B,$ or between $B$ and $D,$ or between $D$ and $A.$
In other words: Just use the most obvious construction of a self-complementary graph of order $4n+1.$
Example: For $G=K_2$ it looks like
