Standard notation for the complete graph with self-loops? $K_n$ is the standard notation for the complete simple (undirected) graph of $n$ vertices, where every vertex is connected to every other. Is there a(n even vaguely) standard notation for the "really" complete (undirected multi)graph of $n$ vertices, where every vertex is connected to every other and to itself - for a total of $\frac{n(n+1)}{2}$ edges, rather than $\frac{n(n-1)}{2}$?
 A: I didn't find anything in Diestel or Bollobas. I don't think there are any "standard" notations. Maybe you could use $\mathring{K_n}$, with $\LaTeX$ as\mathring{K_n}.
A: I'm in with Mike, standard notation is rare in graph theory. People can't even agree if "graph" without additional information refers to "simple graph" (no loops and multiple edges between the same vertices) or "multi graph" (possibly multiple edges between the same vertices; loops or not is again not standard, some call the construct without loops "multi graph", some call it "pseudo graph"), just that it does not mean "digraph" (at most one edge per direction between the same vertices).
The complete graph $K_n$, the cycle graph $C_n$ and the path graph $P_n$ (with $n$ vertices respectively) are more common, but I would always refer once what I mean by these symbols (like I just did), just to be sure. The only case when I would safely omit explanations is when I'm in a lecture and my professor defined exactly what he means by which notation, if I would have to give in exercises I would use his definitions without explanation.
Given your problem, I know of no such notation. Similar to Larry's suggestion, I would define an operation for any graph $G$ to denote the graph resulting from adding loops to any vertex of $G$ and combine this with $K_n$ then.
Since simple graphs can be interpreted as relations ans vice versa, you may want to look up a book about relations, because you are effectively making your relation reflexive with this operation. Maybe there already is a symbol for reflexive closure of relations, I don't know.
