Is there any nontrivial monad on the category of graphs? The question is in the title, but let me specify what I mean by the category of graphs.
In the context of this question, the category of graphs is the category of symmetric irreflexive relations. That means, not the category of symmetric reflexive relations.
That means, no loops and at most one edge between vertices.
 A: I take it morphisms $f: X \to Y$ are by definition functions that preserve the relation: if $x, x'$ are related in $X$, then $f(x), f(x')$ are related in $Y$. 
It's easy to manufacture some silly examples of nontrivial monads, by finding suitable monoidal products on the category of graphs and then finding monoids with respect to that monoidal product. One such monoidal product takes the disjoint sum of two graphs. Then an example of a monoid therein is the one-point graph $1$, which carries a unique monoid structure. 
The associated monad $M$ takes a graph and adjoins an isolated point to the graph, which one could regard as basepoint. If $f: X \to Y$ is a graph morphism, then $M(f)$ is the obvious basepoint preserving extension. To be precise, $M(X) = 1 + X$; the unit of the monad is the inclusion of $X$ in $1 + X$, and the multiplication $1 + 1 + X \to 1 + X$ is the identity on $X$ but identifies the two copies of $1$ as one. 
A: There are billions and billions of them. But it turns out I originally suggested two non-examples:


*

*The non-monad which takes a graph and turns it into the complete graph on the same vertices.

*The comonad which takes a graph and turns it into the discrete graph on the same vertices. (This example was edited after Andreas Blass made his comment.)


And two that still seem to be examples:


*

*The monad which takes a graph and creates a new one with the same vertices, but connects two vertices iff there is a path between them in the original graph.

*The monad arising from $\pi_0$: it takes a graph and returns the discrete graph whose vertices are the connected components of the original graph.

