Let $X$ be a non-reduced Noetherian scheme. We define $K^0(X)$ to be the Grothendieck group of the derived category $Perf(X)$ and $K_0(X)$ to be the Grothendieck group of the derived category $D^b_{coh}(X)$. Then we have a natural Cartan homomorphism $$ K^0(X)\to K_0(X). $$ given by inclusion of the derived categories $Perf(X)\subset D^b_{coh}(X)$.

Next we consider the associated reduced scheme $X_{red}$ with $f: X_{red}\to X$ and we have $K^0(X_{red})$ and $K_0(X_{red})$ also. Moreover there are natural maps $$ f^*:K^0(X)\to K^0(X_{red}) $$ and $$ f_*:K_0(X_{red})\to K_0(X). $$

By the devissage theorem in K-theory, $f_*:K_0(X_{red})\to K_0(X)$ is an isomorphism but this is not true for $f^*$.

My question is: do we have the following commutative diagram $$\begin{array}[c]{ccc} K^0(X)&{\rightarrow}&K_0(X)\\ \downarrow\scriptstyle{f^*}&&\uparrow\scriptstyle{\cong}\\ K^0(X_{red})&{\rightarrow}&K_0(X_{red})? \end{array}$$