homotopy Sym and quotients I'm a novice to homotopical algebra, but I've found myself confronted with it by necessity and have some basic questions... 
I'm going to consider chain complexes over a field $F := \mathbb{F}_2$. Given a chain complex $C$, I'm interested in two operations: 


*

*the ``homotopy Sym'', where I form $(C \otimes C  \otimes E (\mathbb{Z}/2))_{\mathbb{Z}/2}$ where the $\mathbb{Z}/2$ acts diagonally on the tensor product. I'll call this $hSym^2 C$. 

*if $C$ has a $\mathbb{Z}$-action, then I can form ``homotopy quotient'' $C/\mathbb{Z}$, which is $(C \otimes E \mathbb{Z})_{\mathbb{Z}}$, with $\mathbb{Z}$ acting diagonally. I don't know if there is "official" notation for this; I'll just call it $hC/\mathbb{Z}$. 


(Edit: I thought I wrote this down but must have deleted it accidentally; $E G$ is a projective $F[G]$-resolution of the complex $F$ in degree $0$, which is supposed to represent a point; thus $EG$ is morally to be the chain complex of some contractible space on which $G$ acts freely.) 
So my question is about how the composition of these two operations in either order are related. If $C$ has a $\mathbb{Z}$-action, then I think $hSym^2 C$ still has a $\mathbb{Z}$-action, so I could form 
$$
h( hSym^2 C )/\mathbb{Z}
$$
or I could do things in the opposite order: 
$$
hSym^2 (hC/\mathbb{Z}).
$$
Based on naive intuition about how ordinary quotients work, I guess that there should be an induced map 
$$ h( hSym^2 C )/\mathbb{Z} \rightarrow hSym^2 (hC/\mathbb{Z}) $$ 


*

*Is this right? 

*And if it is, then is the above map an (edit:quasi-)isomorphism? (I guess probably not in general)

*How can I understand this map explicitly? For instance, if I choose an explicit model for $E \mathbb{Z}/2$ and $E \mathbb{Z}$, like the standard ones that spit out $\mathbb{RP}^{\infty}$ and $S^1$, then I should in principle be able to write it down explicitly, but I'm confused about how that goes. 

 A: If $C$ has an action of $\mathbb Z$, then $C\otimes C$ has an action of the wreath product ${\mathbb Z}\wr {\mathbb Z}/2$. This is the split group extension $1\to {\mathbb Z}\times{\mathbb Z}\to  {\mathbb Z}\wr {\mathbb Z}/2\to {\mathbb Z}/2\to 1$ associated with the natural action of ${\mathbb Z}/2$ on ${\mathbb Z}\times {\mathbb Z}$. Inside this group you have the subgroup ${\mathbb Z}\times {\mathbb Z}/2$, where ${\mathbb Z}$ is the diagonal subgroup of ${\mathbb Z}\times {\mathbb Z}$.
With this notation, $h(hSym^2C)/{\mathbb Z}$ is the homotopy quotient of $C\otimes C$ by ${\mathbb Z}\times {\mathbb Z}/2$, while $hSym^2(hC/{\mathbb Z})$ is the homotopy quotient of $C\otimes C$ by ${\mathbb Z}\wr {\mathbb Z}/2$. So your map can be understood as the quotient map 
$$[C\otimes C\otimes E({\mathbb Z}\times {\mathbb Z}/2)]_{{\mathbb Z}\times {\mathbb Z}/2}\to [C\otimes C\otimes E({\mathbb Z}\wr {\mathbb Z}/2)]_{{\mathbb Z}\wr {\mathbb Z}/2}$$
associated with the group inclusion ${\mathbb Z}\times {\mathbb Z}/2\hookrightarrow {\mathbb Z}\wr {\mathbb Z}/2$. In particular, the map is not an isomorphism, or even a chain homotopy equivalence, except in the most trivial case.
You can make this map pretty explicit by choosing nice chain level models for $E\mathbb Z$, $E{\mathbb Z}/2$ and $E{\mathbb Z}\wr {\mathbb Z}/2$. As you say, a good model for $E{\mathbb Z}/2$ is given by cellular chains on $S^{\infty}$, with the standard cell structure that has two cells in each dimension. To get a nice model for $E{\mathbb Z}$, think of $\mathbb Z$ acting on $\mathbb R$ by translation. Equip $\mathbb R$ with a cell structure with a zero cell for every integer $n$, and a $1$-cell for every interval $[n, n+1]$. The cellular chain complex gives you a nice small model for $E\mathbb Z$. Since you said you wanted to work with ${\mathbb F}_2$ coefficients, it has the form ${\mathbb F}_2[{\mathbb Z}]\leftarrow {\mathbb F}_2[{\mathbb Z}]\leftarrow 0 \cdots $, where the boundary homomorphism is determined by $[n, n+1]\to [n+1]-[n]$. Once you have models for $E{\mathbb Z}$ and $E{\mathbb Z}/2$, the chain complex $E{\mathbb Z}\otimes E{\mathbb Z}\otimes E{\mathbb Z}/2$ gives a model for ${\mathbb Z}\wr {\mathbb Z}/2$.
