Is an $A$-module homomorphism of the $(A,B)$-bimodule $M$ also $B$-bimodule?

Question

Let $A,B$ be rings, let $M$ be an $A$-module, which is, simultaneously an $A$-module and a $B$-module, and $a(xb)＝(ax)b$ for all $a\in A$, $b\in B$, $x\in M$. Then for an $A$-module homomorphism of $M$, will it must be an $B$-module homomorphism at the same time? Or there are any opposite examples.

Answer 1

Not necessarily: every right $B$-module is a $(\Bbb Z,B)$-bimodule in a unique way. A $\Bbb Z$-linear map $f:M\to N$ of $B$-modules in general won't be a $B$-module homomorphism.

Example: $\Bbb C$ is a $(\Bbb R,\Bbb C)$-bimodule. Complex conjugation is an $\Bbb R$-module map from $\Bbb C$ to $\Bbb C$, but isn't a $\Bbb C$-module map.