TITLE: What is the probability that $\pi(x) + x$ is injective?
QUESTION [11 upvotes]: Let $S$ be a finite group with operator + and $\pi$ be a permutation on $S$.  Then what is the probability that $\pi(x) + x$ is injective over choices of $\pi$?  
The concrete instantiation I'm interested in is $S=$GF$(2^n)$ for fixed $n > 0$.  (Computer Scientists call this "xor on $n$-bit strings.")
For $n=1$ we have two permutations, neither of which produce an injection. For $n=2$ we have 24 permutations, 8 of which induce an injection so the probability is 1/3.
Here is an example $\pi$ for $n=2$:
$$\pi(0)=0,\ \pi(1) = z,\ \pi(z)=z+1,\ \pi(z+1)=1.$$
Here notice that $\pi(x) + x$ produces $0$, $z+1$, $1$, and $z$, respectively, which is an injection.

REPLY [5 votes]: There is a name for what you describe: it's called an orthomorphism. If $S$ is the cyclic group of order $n$, then the number of orthomorphisms for $n=1,3,5,...$ is $1, 3, 15, 133, 2025, 37851,...$ (sequence A006717 in OEIS). (If $n$ is even, $S$ has no orthomorphisms.)
You can find lots of links for the case when $S$ is GF$(2^n)$ by googling "orthomorphism of galois field". For instance, here is a short paper which confirms my $244744192$ and goes on to discuss the construction of permutation polynomials.