Here's my interpretation.
For K
4 the path 0 ⇒ 1 ⇒ 3 ⇒ 2 is a solution because the differences between the vertices of each visited edge are 1, 2, 3 (mod 4), i.e. every natural number x with 0 < x < 4 is present once, and each vertex is visited once. The only other solution is the inverse 0 ⇒ 3 ⇒ 1 ⇒ 2 with differences 3, 2, 1. A path is a solution iff it's inverse is a solution.
K
1 and K
2 both have 1 (trivial) solution, K
3 and higher odd numbers seem to have no solution, K
6 has 4 solutions (diff 1, 4, 3, 2, 5), (diff 2, 5, 3, 1, 4) and their inverses. K
8 has 24 solutions, K
10 has 288 solutions, K
12 has 3856. This sequence is
http://oeis.org/A141599
#lang racket
(require (planet wmfarr/permutations:1:3/permutations))
(define m 12)
(for/fold ((a 0)) ((l (in-permutations (- m 1))))
(define z (make-vector m))
(vector-set! z 0 0)
(for ((i (in-range (- m 1))))
(vector-set! z (+ i 1) (remainder (+ (vector-ref z i) (vector-ref l i) 1) m)))
; (when (permutation? z) (printf "~a ~a~n" l z))
(+ a (if (permutation? z) 1 0)))