Is the moduli space of unorientable Riemann surfaces with $pin^+$ structure orientable? By a non-orientable Riemann surface ${\cal C}$, I mean a compact non-orientable two-manifold without boundary that is endowed with a conformal structure.
Such objects have a moduli space that is somewhat like the more familiar moduli space of oriented Riemann surfaces, but this moduli space is itself non-orientable.  (See corollary 2.3 of https://arxiv.org/abs/1309.0383, where this is proved with a simple explicit example.)
My question is this:  Suppose that ${\cal C}$ is endowed with a $pin^+$ structure.   (This is one of the two possible analogs of a spin structure in the non-orientable case, the other being a $pin^-$ structure.)  I've come to suspect that the moduli space of non-orientable Riemann surfaces with a $pin^+$ structure is itself orientable.  I wonder if this is a known result and where it might be found. 
 A: This is not an answer to the question but rather a comment related to the motivation behind it that might be of interest. A more natural point of view from the perspective of GW theory is to consider symmetric Riemann surfaces i.e. orientable RS with an orientation reversing involution. The moduli of such curves is not orientable but the moduli of equivariant J-hol. maps to a symplectic manifold with an anti-symplectic involution of dimension 2(2n+1) often are and give rise to real GW invariants (in particular this is the case for target curves and CY 3-folds arXiv). The calculations in the target P^1 case suggest that these are related to matrix models for SO/Sp - I hope I'll be able to say something more precise soon. 
A: I think the answer to my question is no, this moduli space is not orientable.  This can be proved via the same example used in corollary 2.3 of https://arxiv.org/abs/1309.0383, which I cited in the question.
A Klein bottle can be constructed as a cylinder whose boundaries are replaced by crosscaps or one-sided circles.   Remove a disc from the Klein bottle; we get a pair of pants or three-holed sphere, except it has one true boundary and two
``boundaries'' that are really one-sided circles.    This is an unorientable manifold with boundary and it can be glued onto the boundary of any possibly orientable Riemann surface with one boundary circle to make a compact example without boundary.
The pair of pants variant described in the last paragraph has a diffeomorphism that acts trivially on the true boundary, exchanges the two one-sided circle ''boundaries'', and preserves the orientation that one would have if one omits
the two one-sided circles.  (This diffeomorphism acts as a Dehn twist by pi near
the true boundary.)    One can pick a pin^+ structure on the pair of pants variant that is preserved by the diffeomorphism.   This diffeomorphism reverses the orientation of the moduli space of conformal structures on the Riemann surfaces
in question, basically because it exchanges the two ''length'' coordinates associated to the one-sided circle ''boundaries.'' 
A: Here is another way of seeing that the pin cover is not orientable. It might provide a helpful perspective when searching for other covers that are orientable.
Two pin structures on a surface are related by a twist in $Hom (\pi_1(X), \pm 1) = H^1(X, \mathbb Z/2)$, and every twist by such a cohomology class gives another pin structure. I think the twist by the orientation class in $H^1(X,\mathbb Z/2)$ just gives the original pin structure, conjugated by a rotation by $\pi$ (which is central in the orthogonal group but not in the pin group).  So the moduli space of pin-structure is a torsor over the moduli space by the group $H^1(X, \mathbb Z/2)$ modulo the orientation class. 
For a genus $g$ surface with a cross-cut added, this group is $(\mathbb Z/2)^{2g}$ and carries a symplectic structure. Moreover, the monodromy of this group over the moduli space is $SP_{2g} (\mathbb Z/2)$, as we can see by removing the cross-cut and using the well-known fact that the monodromy of the moduli space of Riemann surfaces is surjective on cohomology.
This implies that this torsor doesn't cover the orientation torsor, or any other two-torsor. If it did, then the subset of $H^1(X,\mathbb Z/2)$ such that twisting by gives a pin structure that maps to the same branch of the orientation torsor would be monodromy invariant, and contain some nonzero vector, hence contain all nonzero vectors, and so the map would be constant. 
To orient the moduli space you likely need to find a $\mathbb Z/2$-torsor, or a torsor by a group that admits a monodromy-invariant map to $\mathbb Z/2$.
