I'm posting an answer so that this doesn't remain open and unanswered, even though the OP said in the comments he figured it out. I just posted an answer to an analogous question and in doing so found this question. The relevant paper is "Stable model categories are categories of modules" by Schwede and Shipley. The construction the OP describes (Dold-Kan plus shift) is on page 39, where it is also shown how to make this into an $HR$-module (via the Alexander-Whitney map). Remark B.1.10 discusses the symmetric group actions. This functor $\mathcal{H}$ takes a chain complexes to a naive $HR$-module, not a symmetric spectrum. It is not an extension of the usual Eilenberg-MacLane functor, $H$. The authors claim that there is no way to make $\mathcal{H}R$ into a symmetric spectrum that is level equivalent to $HR$. This explains why the Quillen equivalence has to zig-zag through naive $HR$-modules.

I'm posting a CW answer so that this doesn't remain open and unanswered, even though the OP said in the comments he figured it out. I just posted an answer to an analogous question and in doing so found this question. The relevant paper is "Stable model categories are categories of modules" by Schwede and Shipley. The construction the OP describes (Dold-Kan plus shift) is on page 39, where it is also shown how to make this into an $HR$-module (via the Alexander-Whitney map). Remark B.1.10 discusses the symmetric group actions. This functor $\mathcal{H}$ takes a chain complexes to a naive $HR$-module, not a symmetric spectrum. It is not an extension of the usual Eilenberg-MacLane functor, $H$. The authors claim that there is no way to make $\mathcal{H}R$ into a symmetric spectrum that is level equivalent to $HR$. This explains why the Quillen equivalence has to zig-zag through naive $HR$-modules.