Let $M$ be a $\mathrm{II}_{1}$ factor with the faithful trace $\tau$ in standard form. Suppose $\tau'(x')=\tau(Jx'^{*}J)$. If we complete $M$ with $\|\cdot\|_{1}$ norm, i.e., $\|x\|_{1}=\tau(|x|)$, then the completion is $L^{1}(M,\tau)$. My qustion is if we complete $M'$ by $\|x'\|_{1}=\tau'(|x'|)$, is it true that $L^{1}(M',\tau')=L^{1}(M,\tau)$, or they isometrically isomorphic?

Let $M$ be a $\mathrm{II}_{1}$ factor with the faithful trace $\tau$ in standard form. Suppose $\tau'(x')=\tau(Jx'^{*}J)$. If we complete $M$ with $\|\cdot\|_{1}$ norm, i.e., $\|x\|_{1}=\tau(|x|)$, then the completion is $L^{1}(M,\tau)$. My qustion is if we complete $M'$ by $\|x'\|_{1}=\tau'(|x'|)$, is it true that $L^{1}(M',\tau')=L^{1}(M,\tau)$, or they isometrically isomorphic? Well this is clear that it is sometric isomorphism by the comment of Matthew. Now instead of that if we take the vector state $\omega_{\tau}=\langle x\Omega_{\tau},\Omega_{\tau}\rangle$. Complete $M$ and $M'$ with respect to this $\|\cdot\|_{1}$ norm coming from state, are they same? I am confused because this equality $L^{2}(M,\omega_{\tau})=L^{2}(M',\omega_{\tau})$. Thanks in advance!!