I am trying to understand matrix ring spectra. Apparently, I have two different definitions of those and I did not manage to show that they are equivalent - maybe they even are not in the general case.

Given a connective (commutative) symmetric ring spectrum $R$, we define the $k$-th space of $M_n(R)$ to be the following space of pointed maps $$ [M_n(R)]_k = map_*(n_+, n_+ \wedge R_k) $$ where the multiplication is given by $$ n_+ \to n_+ \wedge R_l \to n_+ \wedge R_k \wedge R_l \to n_+ \wedge R_{k+l} $$ i.e. apply the first map, apply the second map wedge identity, then multiply in $R$. The definition is used in this paper to define the algebraic $K$-theory of $R$ by considering $GL_1(M_n(R))$. The latter has the right homotopy type if $R$ is convergent (see the reference). Correct me if I am wrong, but this should be stably equivalent to $End_R(n_+ \wedge R)$, where we consider $n_+ \wedge R$ as an $R$-module spectrum in the obvious way.

My second definition works as follows: Take $R \times \dots \times R$ as an $R$-module with the "diagonal" $R$-action and define $M_n'(R) = End_R(R \times \dots \times R)$. At least if $R$ is semistable, the canonical map $n_+ \wedge R \to R \times \dots \times R$ should be a stable equivalence, but according to the answer to my last question this is not enough to deduce that the endomorphism spectra are stably equivalent.

Does the second definition make sense? Would it be enough to demand that $R$ is fibrant-cofibrant to get the equivalence? Is there a more general condition to achieve equivalence in this situation?

The way I see those two versions is that the space $M_n(R)$ contains matrices with just one non-vanishing column (or row?), whereas $M_n'(R)$ contains full matrices. Going over to homotopy classes of maps from a space into $M_n(R)$ adds the "elementary matrices" together (consider for example a loop running through two different copies of $R$). Thus, up to homotopy it should look like $M_n'(R)$.