Just to lend some context to the above question: the mapping class group of the two-torus is naturally isomorphic to GL(2, Z). If we restrict to orientation preserving homeomorphism the mapping class group is SL(2, Z). The periodic mapping classes (isotopy classes of homeomorphisms) are exactly those with trace less than two in absolute value. (Hmm, and +/- Id, I guess!) Now we need to count the number of conjugacy classes of periodic elements. There should be a cool algebraic way to do this. (Perhaps it would help to give a purely algebraic proof that the order of torsion is at most 6?)

I think that there is a geometric way to do this: every periodic element occurs as the symmetry of some flat torus (= parallelogram with opposite sides identified). All tori have have the hyperelliptic symmetry, corresponding to rotation by 180 degrees about any point. (These maps lie in the mapping class of the negative identity.) Other symmetries:

Rombic tori have a reflection symmetry as do rectangular tori.

The square torus has a rotation by 90 degrees.

The hexagonal torus has a rotation by 60 degrees.

So I count:

1. the identity, Id

2. the hyperelliptic = -Id = rotation by 180

3. rotation by 90

4. rotation by 60

5. rotation by 120

6. the reflection [[-1,0],[0,1]] (reflection in an axis) and

7. the reflection [[0,1],[1,0]] (exchange axes).

You can prove that the last two are distinct algebraically. Perhaps the lack of 45 degree rotation is a geometric proof.

Now, we could perform similar geometric tricks to obtain symmetries of $N_4$ and get at least all of the rotations... [Edit: For example, it is possible to build a copy of $\rm{Sym}_4$ by placing the cross-caps at the vertices of a tetrahedron.]