I felt that the earlier question may be too challenging, so let me provide a different angle and more infos to tackle an easier and separate problem.
Let us consider a more explicit a short exact sequence: $$ 1\to SU(n)\to U(n)\to U(1) \to 1, $$ where the normal subgroup $N=SU(n)$, $Q=U(1)$, and $G=U(N)$ can be continuous Lie groups in general (or finite groups).
We see that $ {U} (n)$ is related to $ {SU} (n)$ and $ {U} (1)$ by $$ {U} (n)=\frac{ {U} (1)\times {SU} (n)}{\mathbb{Z}_n} $$ where ${\mathbb{Z}_n}= {\mathbb{Z}}/({n \mathbb{Z}})$, a finite Abelian cyclic group of order $n$. This is the case because both $U(1)$ and ${SU} (n)$ shares the same subgroup $${\mathbb{Z}_n}=\{ \exp(\frac{2 \pi i }{n}j) \cdot \mathbb{I}_{n\times n}\}$$ where $j \in \mathbb{Z} \mod n$. Let us take $N \in$ even integer, thus $${\mathbb{Z}_n} \supset \{1,-1\}={\mathbb{Z}_2}.$$
Say we already get the data and the computations of the bordism group $$ \Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2}=\Omega_{d}^{(Spin \times U(1))/\mathbb{Z}_2}=\Omega_{d}^{Spin_c}, $$ where ${(Spin \times Q)/\mathbb{Z}_2}$ means the modification of the $Spin$-structure to a new ${(Spin \times Q)/\mathbb{Z}_2}$-structure. Here the $Spin$ and $Q$ shares a normal subgroup $\mathbb{Z}_2$ that was mod out, such that $$ Spin/\mathbb{Z}_2= SO, $$ or more explicitly $$ Spin(d)/\mathbb{Z}_2= SO(d), $$ where we omit the dimension $d$ through this post in the $Spin \equiv Spin(d)$.
Suppose we also know the data and the computations of the bordism group $$ \Omega_{d}^{(Spin \times N)/\mathbb{Z}_2}=\Omega_{d}^{(Spin \times SU(n))/\mathbb{Z}_2}, $$ where ${(Spin \times N)/\mathbb{Z}_2}$ means the modification of the $Spin$-structure to a new ${(Spin \times N)/\mathbb{Z}_2}$-structure. Here the $Spin$ and $N$ shares a normal subgroup $\mathbb{Z}_2$ that was mod out.
Question How can we compute $$\Omega_{d}^{(Spin \times G)/\mathbb{Z}_2}=\Omega_{d}^{(Spin \times U(n))/\mathbb{Z}_2}=? $$ where the precise extension is given, based on the previously known information of $\Omega_{d}^{(Spin \times Q)/\mathbb{Z}_2}=\Omega_{d}^{Spin_c}$ and $\Omega_{d}^{(Spin \times N)/\mathbb{Z}_2}=\Omega_{d}^{(Spin \times SU(n))/\mathbb{Z}_2}$?
For example, we know the $\Omega_{d}^{Spin_c}$ data as follows: $$\Omega_{0}^{Spin_c}=\mathbb{Z},$$ $$\Omega_{1}^{Spin_c}=0,$$ $$\Omega_{2}^{Spin_c}=\mathbb{Z},$$ $$\Omega_{3}^{Spin_c}=0,$$ $$\Omega_{4}^{Spin_c}=\mathbb{Z}^2,$$ and let us say we can take some specific $N \in$ even say as $N=4$. $$\Omega_{0}^{(Spin \times SU(4))/\mathbb{Z}_2}=\mathbb{Z},$$ $$\Omega_{1}^{(Spin \times SU(4))/\mathbb{Z}_2}=\mathbb{Z}_2,$$ $$\Omega_{2}^{(Spin \times SU(4))/\mathbb{Z}_2}=\mathbb{Z}_2,$$ $$\Omega_{3}^{(Spin \times SU(4))/\mathbb{Z}_2}=0,$$ $$\Omega_{4}^{(Spin \times SU(4))/\mathbb{Z}_2}=\mathbb{Z}^2,$$ How do we determine
$$\Omega_{d}^{(Spin \times G)/\mathbb{Z}_2}=\Omega_{d}^{(Spin \times U(n))/\mathbb{Z}_2}=\Omega_{d}^{(Spin \times U(4))/\mathbb{Z}_2}=? $$
How about other even $n$?
