spin mapping class group of circles The MCG of the circle is $\mathbb{Z}/2$, generated by an orientation-reversing diffeomorphism.
The circle has two spin structures, a periodic one and an anti-periodic one. Each has a nontrivial orientation-preserving spin diffeomorphism that exchanges the two sheets of the spin structure.
The spin-MCG -- defined to be the group of isotopy classes of spin diffeomorphisms -- would seem to be $\mathbb{Z}/2\times\mathbb{Z}/2$ for both the periodic and anti-periodic circles. Is this correct?
 A: A spin mapping class of a spin manifold $(M, \mathfrak s)$ is a choice of oriented diffeomorphism $f: M \to M$ and isomorphism $f^* \mathfrak s \cong \mathfrak s$, modulo isotopy through such. These two pieces of data are somewhat orthogonal. The first thing to determine is if a map $f$ indeed has $f^* \mathfrak s \cong \mathfrak s$ at all before thinking about choices of such an isomorphism. 
We have a short exact sequence of groups of the form 
$$0 \to \text{Aut}(\mathfrak s) \to \text{Diff}^{\text{spin}}(M) \to \text{Diff}^{\text{spinnable}}(M) \to 0,$$ where the first term denotes the group of bundle automorphisms of $\mathfrak s$ (isomorphic to $\Bbb Z/2$, the 'swap' action) and the last term is the subgroup of $\text{Diff}^+(M)$ of oriented mapping classes that admit some isomorphism $f^*\mathfrak s \cong \mathfrak s$. Taking homotopy groups we obtain part of a long sequence $$\pi_1 \text{Diff}^{\text{spinnable}}(M) \to \Bbb Z/2 \to \text{MCG}^\text{spin}(M) \to \text{MCG}^{\text{spinnable}}(M) \to 0.$$ 
Now for $M = S^1$, the group $\text{Diff}^{\text{spinnable}} = \text{Diff}^+$, as every oriented diffeomorphism is isotopic to the identity, and hence $f^* \mathfrak s \cong \text{Id}^* \mathfrak s = \mathfrak s$. The issue is determining the boundary map $\Bbb Z = \pi_1 \text{Diff}^{\text{spinnable}}(S^1) \to \Bbb Z/2$. 
Let us compute this explicitly; our loop downstairs will be $S^1 \to \text{Diff}^{\text{spinnable}}(S^1)$ acting by rotation; we will lift this to a path of spin diffeomorphisms. In 1D, a spin structure is the same thing as a choice of double cover of the circle. In the periodic spin structure (the one with disconnected double cover, I think this term means), it is easy to lift this loop of diffeomorphisms to a loop of diffeomorphisms of the double cover: just do the corresponding rotations on both components. 
For the antiperiodic spin structure, we write the double cover as $S^1 \to S^1$, given by $z \mapsto z^2$. The natural diffeomorphism to lift rotation by $\theta$ to upstairs is the rotation by the angle $\theta/2$; taking $\theta \to 2\pi$, we find that while the diffeomorphism downstairs is trivial (it is the endpoint of a loop), the diffeomorphism upstairs is not (it swaps the fibers). That is, in the antiperiodic case, the map $\pi_1 \text{Diff}^{\text{spinnable}}(S^1) \to \Bbb Z/2 = \text{Aut}(\mathfrak s_a)$ is nontrivial.
Thus in the case of the periodic spin structure one has $\text{MCG}^{\text{spin}}(S^1; \mathfrak s_p) = \Bbb Z/2$ and in the case of the antiperiodic spin structure one has $\text{MCG}^{\text{spin}}(S^1; \mathfrak s_a) = 0.$
It seems to me the reason you get a different answer is that you are trying to include the reflection as a generator, but it doesn't make sense to lift that to "preserving a spin structure" when it already doesn't preserve orientation; you could at most see that it preserves a Pin structure. 
