Two rings A and B are isomorphic. Then $End_A(M) \cong End_B(M)$ for every A and B Modul M.

Question

Hi I found this Lemma without a proof in a script: If two rings A and B are isomorphic with $\Phi A \to B$. And M is an B-Leftmodul then M is also an A-Leftmodul with with $A \times M \to M: am \mapsto \Phi(a)m $. Then $End_A(M) \cong End_B(M)$ as Rings. The first part is clear but why are the endomorphism rings isomorphic? My guess what be that the Isomorphism is the identity and for $a\in A, f \in End_B(M), m\in M, f(am)=f(\Phi(a)m)=\Phi(a)f(m)=af(m)$. So these rings are identical. Is that correct?