Does anyone know what is $KO^{tC_2}[1/2]$? (with the trivial action)

@DenisNardin I think I got a proof that $π_*KO^{tC_2}[1/2]=\mathbb{Q}_2[v_1^{\pm2}]$, but confirmations are welcome...

@DenisNardin I agree with your calculation. And you don't need to invert 2, already \pi_* KO^{t C_2} = Q_2[t^{\pm 1}]. (Where t comes from a Thom class for 4*sign.)