I'm not sure this qualifies, but here goes.

It doesn't take any fancy enumerations to show that the average number of fixed points is 1: Among the $n!$ permutations of $n$ objects, each object is fixed by $(n-1)!$ permutations, for a total of $n!$ fixed points. So if the probability of having no fixed points tends to 0 (for some subsequence of $n$'s), then the probability of having more than one fixed point must also tend to 0. This seems like it should be easily disprovable nonsense.

Added 1/20/12 Brendan McKay's "switching method" answer nails the nonsense. Here's a slight variation on his argument.

Assume you're at a large $n$ where, say 99 percent (or more) of the permutations have exactly one fixed point. (That was the thrust of my intial posting: If the fraction of permutations with no fixed points is close to 0, then the fraction with exactly one fixed point must be close to 1.) Then there's at least one object that's fixed by at least $.99(n-1)!$ permutations having no other fixed points. If you take each of these permutations and switch the fixed object with each of the other $n-1$ objects, you've created $.99(n-1)!(n-1) = .99n!(1-1/n)$ distinct permutations with no fixed points. Since $n$ is large, this is almost $.99n!$ itself, so certainly greater than $.5n!$. But $.99+.5 > 1$, which gives us more than $n!$ permutations on $n$ objects.

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I'm not sure this qualifies, but here goes.

It doesn't take any fancy enumerations to show that the average number of fixed points is 1: Among the $n!$ permutations of $n$ objects, each object is fixed by $(n-1)!$ permutations, for a total of $n!$ fixed points. So if the probability of having no fixed points tends to 0 (for some subsequence of $n$'s), then the probability of having more than one fixed point must also tend to 0. This seems like it should be easily disprovable nonsense.