The computation of the unoriented bordism group of the point $N\_\*=\Omega\_\*^O$ is a classic result.


I would like to know a related bordism group, where we specify the twisted fundamental class $[M]\in H\_d(M,\mathbb{Z}^w)$ as part of the data. More precisely, I would like to consider the pairs 
$$
(M, [M]\in H\_d(M,\mathbb{Z}^w))
$$ where $\mathbb{Z}^w$ is the coefficient system twisted by $w\_1(M)$; I then call two $(M,[M])$ and $(M',[M'])$ bordant when there is $(N,[N]')$ such that $\partial N=M \sqcup M'$ where the twisted orientation of $N$ induces that of $M$ and $M'$, as in the case of oriented bordism.


Most probably this is well known to the experts...



Unoriented bordism with twisted orientation
I think this theory is the same as unoriented bordism, when one tries to make sense of it. 


As it stands I don't think it makes sense, because I think that the expression "$\mathbb{Z}^w$" does not describe a local system on $M$, but rather an isomorphism class of local systems, and hence $H\_d(M ; \mathbb{Z}^w)$ is an isomorphism class of abelian group, and hence it is not meaningful to talk about an element of it.


A manifold $M$ has an "orientation" local system(=locally constant sheaf). This is the locally constant sheaf with values $\mathcal{O}\_M(U) = H\_d(M, M \setminus U; \mathbb{Z})$ on all balls $U \cong \mathbb{R}^d$ in $M$. I am sure that $\mathcal{O}\_M$ is isomorphic to whatever might be meant by $\mathbb{Z}^w$, but $\mathcal{O}\_M$ is an actual local system.


Any $d$-dimensional closed manifold $M$ has a unique fundamental class $[M] \in H\_d(M ; \mathcal{O}\_M)$, where $\mathcal{O}\_M$ is the "orientation" local system of $M$, and fundamental class means that it restricts to the canonical generator of
$$H\_d(\mathbb{R}^d, \mathbb{R}^d\setminus \{0\} ; \mathcal{O}\_{\mathbb{R}^d}) \cong\_{UCT} H\_d(\mathbb{R}^d, \mathbb{R}^d\setminus \{0\} ; \mathbb{Z}) \otimes H\_d(\mathbb{R}^d, \mathbb{R}^d\setminus \{0\} ; \mathbb{Z})$$
for every ball $\mathbb{R}^d \cong U \subset M$. Note that this rank 1 $\mathbb{Z}$-module indeed has a preferred generator as it is the tensor square of a rank 1 $\mathbb{Z}$-module. That such a fundamental class is unique if it exists, and that it exists, is by the usual proof of existence of fundamental classes.


Made precise in this way the question is vacuous, because $[M]$ is seen to not be a choice of data.


