The following doubt came after reading the book "Hilbert C*-modules" by E.C. Lance. Let $A$ be a C*-algebra and $E$ a Hilbert $A$-module, there's a natural structure of Hilbert $A$-module on $E^n$ given by $$\langle (x_1,\dots,x_n),(y_1,\dots,y_n)\rangle_A = \sum \langle x_i,y_i\rangle$$ but also on page 39 of the book we're introduced a Hilbert $M_n(A)$-module structure on $E^n$ given by $$(x_1,\dots,x_n)\cdot (a_{ij})=\left(\sum x_ia_{i1},\dots ,\sum x_ia_{in}\right)$$ and $$\langle (x_1,\dots,x_n),(y_1,\dots,y_n)\rangle_{M_n(A)}=(\langle x_i,y_j\rangle)$$ Later on, on page 58, there's a result that states $\mathcal{L}_{M_n(A)}(E^n)\simeq \mathcal{L}_{A}(E^n) $. The *-homomorphism that establishes this isomorphism according to results on the previous pages seems to be $T\mapsto T$. I'm pretty sure this isn't exactly the isomorphism since $T$ being adjointable in the $M_n(A)$ sense doesn't seem to imply it being adjointable in the $A$ sense.
My question is: explicitly what would be the isomorphism between these two algebras? If $\varphi:\mathcal{L}_{M_n(A)}(E^n)\rightarrow \mathcal{L}_{A}(E^n)$ is the isomorphism then what would $\varphi(T)(x_1,\dots,x_n)$ be?
Realized the morphism $T\mapsto T$ actually works. Take $x,y\in E^n$, $T\in \mathcal{L}_{M_n(A)}(E^n)$ and write $T=(T_1,\dots,T_n)$ and $T^*=(S_1,\dots,S_n)$ where $T_i,S_i:E^n\rightarrow E$ are linear maps. Then $$(\langle T_i(x),y_j \rangle)_{i,j=1}^n=\langle T(x),y \rangle_{M_n(A)}=\langle x , S(y) \rangle_{M_n(A)}=(\langle x_i,S_j(y) \rangle)_{i,j=1}^n$$ meaning $\langle T_i(x),y_j \rangle= \langle x_i,S_j(y) \rangle$ for any $i,j\in \{1,\dots, n \}$. But now $$\langle T(x),y \rangle_{A}=\sum_{i=1}^n \langle T_i(x), y_i\rangle=\sum_{i=1}^n \langle x_i, S(y)\rangle=\langle x , T^*(y) \rangle$$ this means not only that $T\in \mathcal{L}_A(E^n)$ but that it has the same adjoint as in the $M_n(A)$ case and therefore $T\mapsto T$ is a *-isomorphism between $\mathcal{L}_{M_n(A)}(E^n)$ and $\mathcal{L}_A(E^n)$.