Certain isometry overfinite ring is product of isometries over each local factor

Let $R$ be a finite ring, $R_1 \times \dots \times R_m$ its decomposition into finite local rings. Let $\pi_i$ denote the projection to each factor $R_i$. Then $\pi_i$ extends componentwise to a map $\pi_i: \operatorname{Mat}_{m \times n}(R) \to \operatorname{Mat}_{m \times n}(R_i)$, and, for $C$ an $R$-module, we have $C \cong \pi_1(C) \times \dots \times \pi_m(C)$, each $\pi_i(C)$ is an $R_i$-module, and for the $R$-length of $C, \lambda_R(C) = \sum_{i=1}^m \lambda_{R_i}(\pi_i(C))$. Let $\phi: \operatorname{Mat}_{m \times n}(R) \to \operatorname{Mat}_{m \times n}(R)$ be a linear map such that $\operatorname{length}_R(\operatorname{rowsp}(\phi(A))= \operatorname{length}_R(\operatorname{rowsp}(A))$, for each matrix $A$, where $\operatorname{rowsp}(A)$ denotes the $R$-module generated by the rows of $A$. It can be proven that, when $R$ is a local ring, a length-preserving map as described above is just $\phi(A) = PAQ$, for all matrices $A$, where $P,Q$ are suitable invertible matrices, up to transposition of $A$ in the square case. My question is, letting $A$ decompose into $A_1 , \dots , A_m$ following the above, is it true that every $\phi$ as above decomposes into $\phi_1 \times \dots \times \phi_m$, where each $\phi_i$ is a map that preserves the length (as an $R_i$- module) of the rowspace of $A_i$? This feels very intuitive to me, but somehow I can't see how to write a rigorous argument.

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