A function f is an automorphism of A iff f is an isomorphism of A onto
itself. A structure A is rigid iff the only automorphism of A is the
identity function from A onto A.  Let G_n be the set of simple graphs
with node set {1,...,n} and let G = (union of all finite n) G_n.  Let
R contained in G be the set of ridig simple graphs and let R_n = R
(intersect) G_n. The problem is to show that lim (n->infinity) |R_n| /
|G_n| = 1.

Desired time is one week.

I believe the answer is yes and should exist in the standard literature.
The question is not completely unambiguous, but I could assist you in
clarifying and finding a satisfactory answer within the time frame
that you indicate.

Unfortunately though, since I am not a google researcher, I am unable
to be paid for this
effort and am unwilling to perform this service voluntarily.