I am a beginner in algebraic K-theory and I want to make sure that I understand the following correctly: Let $f:A \to A'$ be an isomorphism of commutative rings. Denote by $\mathcal{P}(A)$ (resp. $\mathcal{P}(A')$) the exact category of finitely generated projective $A$ (resp. $A'$) modules. This induces a natural transformation $\mathcal{P}(A) \to \mathcal{P}(A')$ which simply takes $M$ to $M$, the second $M$ considered as $A'$ module (since $A' \cong A$). Is this correct?
If this is correct this should mean that given two isomorphisms $f,g$ from $A$ to $A'$ the induced morphisms on the Quillen K-groups (recall $K_i(\mathcal{P}(A))=\pi_{i+1}(BQ\mathcal{P}(A),o)$ where $o$ is the zero object in the category $\mathcal{P}(A)$, similarly for $A'$), $K_i(f), K_i(g)$ from $K_i(\mathcal{P}(A))$ to $K_i(\mathcal{P}(A'))$ are identical. Here I am using Lemma $3.6$, page 40 from the book "Algebraic K-theory" by V. Srinivas which states that given a natural transformation between the two functors from $\mathcal{P}(A)$ to $\mathcal{P}(A')$ (induced by $f$ and $g$), the induced functors from $Q\mathcal{P}(A)$ to $Q\mathcal{P}(A')$ are homotopic. Is this correct i.e., is $K_i(f)=K_i(g)$ for all $i \ge 0$?
Beware that although the functors $A\text{-proj}\to A^\prime\text{-proj}$ defined by $f$ and $g$ are both the identity functors on the level of underlying abelian groups, they are not isomorphic - even more, there's not even a canonical choice of natural transformation between them to which you could apply Lemma 3.6 from Srinivas' book.
In fact, $K_i(f)$ and $K_i(g)$ need not coincide. For example, take $A=A^\prime=k\times k$ for some field $k$, so that $A\text{-proj}\cong k\text{-vect}\times k\text{-vect}$ and hence $K_0(A)={\mathbb Z}\times{\mathbb Z}$. Then, the automorphism $g: A\to A$ flipping the two $k$ factors induces the flipping map on ${\mathbb Z}\times{\mathbb Z}$, which is different from the identity map induced by the identity automorphism $f=\text{id}_A$.