Are these two definitions of $EG$ equivalent? Let $G$ be a topological group with multiplication $\sigma:G\times G\to G$.
The simplicial topological space $\mathcal{E}G$ defined by
$$
\ldots \begin{array}{c}\to\\\to\\\to\\\to\end{array}G\times G\times G\begin{array}{c}\xrightarrow{(\sigma,id)}\\\xrightarrow{(id,\sigma)}\\\xrightarrow{pr_{0,1}}\end{array}G\times G \begin{array}{c}\xrightarrow{\sigma}\\\xrightarrow{pr_0}\end{array}G
$$
admits a $G$-action with is in degree $n$ induced by 
\begin{eqnarray}
G\times \mathcal{E}G_n&\to& \mathcal{E}G_n\\
(g,(g_0,\ldots,g_n))&\mapsto&(gg_0,g_1,\ldots,g_n)
\end{eqnarray}
and the quotient of $\mathcal{E}G$ by this action defines $\mathcal{B}G$. The realization $EG\to BG$ of the quotient map $\mathcal{E}G\to \mathcal{B}G$ is the construction of the universal $G$-bundle, at least this is what I remember from my topology course.
Now I came across an other definition for the simplicial space $\mathcal{E}G$ which I will temporarily call $\mathcal {E}'G$. It is given by
$$
\ldots \begin{array}{c}\to\\\to\\\to\\\to\end{array}G\times G\times G\begin{array}{c}\xrightarrow{pr_{1,2}}\\\xrightarrow{pr_{0,2}}\\\xrightarrow{pr_{0,1}}\end{array}G\times G \begin{array}{c}\xrightarrow{pr_0}\\\xrightarrow{pr_1}\end{array}G
$$
and the diagnoals as degenerecies. It seems that $G$ acts on this space diagonally, i.e.
\begin{eqnarray}
G\times \mathcal{E}'G_n&\to& \mathcal{E}'G_n\\
(g,(g_0,\ldots,g_n))&\mapsto&(gg_0,gg_1,\ldots,gg_n)
\end{eqnarray}
and the quotient should define $\mathcal{B}'G$.
My question is how $\mathcal{E}G$ and $\mathcal{E}'G$ are related? Is there a levelwise homotopy equivalence or isomorphism which respects the $G$-action (i.e. is equivariant)? Does this imply that $B'G$ and $BG$ are ''the same''?
 A: The two are isomorphic: the $E'G \to EG$ map is the easier written down of the two, given levelwise as $(g_1,g_2,\dots, g_n) \mapsto (g_1, g_1^{-1}g_2, g_2^{-1}g_3, \dots, g_{n-1}^{-1}g_n)$. The map in the other direction $EG \to E'G$ takes $(h_1, \dots, h_n) \mapsto (h_1, h_1h_2, h_1h_2h_3, \dots, h_1h_2h_3\dots h_n)$.(Sort of a "partial products" map)
You can see there's really no other choice for this. The degeneracies in $E'G$ are from the diagonal map as you say, and those in $EG$ must take the 0-simplex $g$ to the n-simplex $(g,*,*,\dots,*)$.
Note that the construction for $E'G$ you've written can be written down for any set $X$, while the one you wrote for $EG$ uses the group law. This has the effect of making the group action more obvious for $EG$ than $E'G$, but they both give the same $BG$. 
Digression if G is discrete:
One way to think of this is that $BG$ is the nerve of the category $C$ with one object *, and $\hom(*,*) = G$. 
It can be thought of as a quotient of the groupoid $D'$ with objects indexed by $G$ with $\hom(g,h) = *$, and $ND$ corresponds with $E'G$. The category $D'$ is contractible because every object is a terminal (and final, for that matter) object. (And, again, you can do this for any set of objects)
Or you can think of it as a quotient of the category $D$ that is the comma category $C \downarrow *$ of objects over * in $C$, which is readily seen to be contractible as it has a terminal object $id : * \to *$ (actually, again every object is terminal). Here $ND$ is more readily seen to be the same as $EG$. (And, again, to construct this category at all, you need to keep in mind morphism composition = group multiplication)
