The pair of species with the same bond order is:

Calculating bond orders:${\text{N}}_2$ (14e) $=3.0$; ${\text{O}}_2$ (16e) $=2.0$.${\text{O}}_2^{2-}$ (18e): B.O. $=\frac{10-8}{2}=1.0$.${\text{B}}_2$ (10e): Config ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}({\pi }_{2p_x}^1={\pi }_{2p_y}^1)$. B.O. $=\frac{6-4}{2}=1.0$.Both have B.O. of 1.0.${\text{O}}_2^{+}$ (15e) $=2.5$; ${\text{NO}}^{+}$ (14e) $=3.0$.$\text{NO}$ (15e) $=2.5$; $\text{CO}$ (14e) $=3.0$.