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Alternatively one can take the simplicial $G$-space $G^{\cdot+1}$ all of whose face maps are projection maps and whose degeneracies are diagonal maps. You can then take the product of this simplicial $G$-space with the constant simplicial $G$-space $X$. To obtain a simplicial $G$-space which levelwise looks the same as the one in the previous comment, but with different structure maps and different $G$-actions (this one is diagonal and realizes the Borel construction). When $X$ is $G$-free the augmentation admits an extra degeneracy and the the associated cochain complex is exact.
I don't think the projection maps will define a simplicial diagram. For example, if your diagram was simplicial, there should be an equality between the two composite face maps $(g_1,g_2,x)\mapsto g_1 g_2 x$ and $(g_1,g_2,x)\mapsto g_1 x.$ The standard construction uses the group multiplication on $G$ not the projection. Using this construction one obtains an augmented simplicial $G$ space $G^{\cdot+1}\times X\rightarrow X$. This is the bar construction on $X$ and the geometric realization is $G$-homeomorphic to the Borel construction.
@Prasit: I did not fully understand your comment, but I will reiterate how my answer relates to your question. The case in your question is $MSL_1(S_p)$ whose $\pi_0$ is torsion-free. Since I suspected you meant to ask about $MGL_1(S_p)$ I included that case. That spectrum is universal in the sense that it is contractible and hence terminal. I am claiming that the homotopy groups of $MG$ are $\bZ/p$-modules and calculating them is equivalent to calculating the homology as as a comodule over the dual Steenrod algebra. I did not claim that the latter problem was easy; it is just algebraic.
@Prasit: $R$ is a wedge of suspensions $H\mathbb{Z}/p$'s so you get one $\mathbb{Z}/p$ in the homotopy groups for each summand and for each such summand you get a copy of the dual Steenrod algebra in the homology.
. whose composite is multiplication by $|G|$ and which factors through $\mathrm{Ho}(M)^G(X,Y)$. From here one can see that $\mathrm{Ho}(M^G)(X,Y)\cong\mathrm{Ho}(M)^G(X,Y)$ when $|G|$ acts invertibly. It appears the non-trivial part is seeing that $\mathrm{tr}$ descends to the homotopy category.
As often happens, behind a vanishing spectral sequence argument lies a more elementary argument. My more involved method is probably overkill in this case. Nonetheless, the argument indicates why it is true. For $X,Y\in M^G$ we have a map $$ \mathrm{tr}\colon M(X,Y)\rightarrow M^G(X,Y)$$ given by $f(-)\mapsto \sum_{g\in G} (g\cdot f)(-)=\sum_{g\in G} gf(g^{-1} -)$. I suspect that $\mathrm{tr}$ descends to homotopy categories, perhaps by an argument with right homotopies. If this is the case, then precomposing with the restriction functor gives a self map on mapping sets in $\mathrm{Ho}(M^G)$..
