Quaternary Symmetry
There are just five closed knight tours with quaternary symmetry on the 6×6 board. They were given by Paul de Hijo (1882) [and possibly earlier by Carl Adam (1867) but I have not yet been able to check this reference] and these most attractive tours have been independently rediscovered many times since, e.g. by Bergholt (1918) and Cozens (1940). The central angles of the 5 quaternary tours are in our lettering a, e, g, i, l, all four angles in a quaternary tour being of the same type of course. The 'a' tour uses 12 slants (8 provided by the 'a's, and 4 others) but the 'e', 'g', 'i' and 'l' tours use only 4 slants (in these, each central angle provides one slant).
Probably the simplest method of finding all the tours of this type is the graphical one of using blank diagrams and drawing in all the possible patterns of angles on the four central cells, then step by step filling in all possible quaternary linkages between these given moves and the vacant cells. For example the impossible case of four f-angles is illustrated below. First put in the f-angles and the moves through the corners, which are common to all closed tours. Then the moves through the next-to-corner cells a5, b1, e6, f2 are fixed (since moves like a5-c4 are blocked). Then the moves through the edge cells a4, c1, d6, f3 are fixed (since moves like a4-c3 and a4-c5 are blocked). We now have two eight-move short circuits (a2-c3-b5-d6-f5-d4-e2-c1 is one), showing that a tour with these centre angles is impossible. We can however join up the loose ends (a3-c2 etc) to form a quaternary pseudotour of three component paths. The c-angles similarly produce a quaternary pseudotour of five circuits.
A method that may be considered more 'systematic' or mechanical is to use the octonary numbering of the board, and the list of permissible knight-move transitions as shown above. The procedure is to list all possible sequences of cell numbers, starting say from 0, seeking a sequence in which each diagonal number (0, 2, 5) occurs once and each off-diagonal number (1, 3, 4) twice. If one arrives at a sequence of less than nine numbers which cannot be extended because the numbers have already been used one deletes that sequence.
To shorten the procedure one can also take account in advance of other deducible features; such as that the transition 11 cannot occur in oblique quaternary symmetry (since by itself it forms a four-move circuit), and that if an off-diagonal number is repeated one of the numbers between the two occurrences must be a diagonal number (for example 343 represents a switchback, but 323 is two moves across the diagonal; and sequences like 3413, 1431, 4134 cannot occur since they prevent the corner moves 151 which must occur).
Beginning at the centre (0) one ends up with the ten sequences: 0151432340, 0234151340, 0234151430, 0323415140, 0341514320, 0415132340, 0415143230, 0431514320, 0432315140, 0432341510 (where the 0 has been repeated at the end to indicate closure). These occur in pairs that are reversals of each other. This duplication serves as a check on the correctness of the count. This type of method, with different numberings, was used by de Hijo (1882) and Bergholt (1918) in their solutions of this problem.
Binary Symmetry
There are a further 17 closed knight tours with binary symmetry (necessarily rotational symmetry of Eulerian type) on the 6×6 board. Euler (1759) gave a single example. The complete set was first given by Kraitchik (1927) [or possibly much earlier by Carl Adam (1867)]. The central angles of the 17 binary tours are a/b, a/l, b/c, b/i, b/n, c/d, c/f, c/g (3 times), c/l, e/g, f/g, f/i, g/i, g/n, i/l, two angles being of one type and two of another.
No case is possible in which all four angles are alike; tours with angles all alike turn out to be either quaternary or asymmetric. Using the graphical approach it is quite simple to check this enumeration of the tours that have binary symmetry. In terms of number of slants there are: 1 with 12 slants, 10 with 8 slants and 6 with 4 slants.