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\begin{document}

The proof that the equivariant linear isometries operad is an $E_{\infty}$ 
$G$-operad is parallel to the nonequivariant proof, but you are right that
I seem never to have written it up. It belongs in the infinite loop space
paper.  Let $\LA$ be a subgroup of $G\times \SI_n$. Since $\SI_n$ acts 
freely, there can be no $\LA$-fixed point if $\LA\cap \SI_n$ has a point
other than $e$.  Let $\LA\cap \SI_n$ be trivial. Then there is a subgroup
$H$ of $G$ and a homomorphism $\rh\colon H\rtarr \SI_n$ such that $\LA$ is 
the set of points $(h,\rh(h))$. We may view $\sL(n)$ as an $H$-space by
pullback along this isomorphism from $H$ to $\LA$. So it suffices to prove
that $\sL(n)$ is $H$-contractible under this pullback action. At this point,
$\SI_n$ is irrelevant, and we can replace $G$ by $H$ and consider a finite
or countable dimensional $G$-representation $V$, a complete $G$-universe
$U$, and the $G$-space $E=\sI(V,U)$. It suffices to prove that $E^G$ is
contractible, and it is certainly non-empty since $V$ is $G$-isomorphic
to a subrepresentation of $U$.

We mimic the proof of I.1.3 in $E_{\infty}$ ring spaces and $E_{\infty}$
ring spectra. Write $U$ as the direct sum $\sum_{\rh} V_{\rh}^{\infty}$, where the $V_{\rh}$ 
run through the irreducible representations $\rh$ of $G$. We divide $V_{\rh}^{\infty}$
as the direct sum of the even and the odd numbered copies of $V_{\rh}$. Define
$\al\colon U\rtarr U$ to be the obvious $G$-linear isometry mapping the $i^{\text{th}}$ 
copy of each $V_{\rh}$ to the $2i^{\text{th}}$ copy. Define a $G$-linear isomorphism
$\be\colon U\rtarr U\oplus U$ by mapping the $2i^{\text{th}}$ copy of $V_{\rh}$ to the 
$i^{\text{th}}$ copy in the first summand and the $(2i-1)^{\text{st}}$ copy to the $i^{\text{th}}$ copy in the 
second summand. Then $\be\com \al = i_1$, the inclusion of the first summand.
There is a path $H_1$ in $\sI(U,U)^G$ from the identity to $\al$ and a path
$H_2$ in $\sI(V,V\oplus V)^G$ from $i_1$ to $i_2$. They are obtained by normalizing 
the $G$-linear paths $J_1$ and $J_2$ given by
\[  J_1(t)(v) = (1-t)v + t\al(v) \ \ \ \text{and} \ \ \ J_2(v) = (1-t)i_1(v) + ti_2 (v) \]
For each $t$, these are $G$-linear inclusions; that is $J_1(t)(v) = 0$ implies $v=0$ and
similarly for $J_2$. Choose any fixed $\ga\in \sI(V,U)^G$ and define a homotopy
\[ H\colon I\times \sI(V,U)^G \rtarr \sI(V,U)^G \]
by
\[ H(t,f) =  \left\{ \begin{array}{ll}
H_1(2t)\com f   &  \mbox{if  $0\leq t\leq 1/2$} \\
           \be^{-1}\com (f\oplus \ga)\com H_2(2t-1) & \mbox{if $1/2\leq t\leq 1$.} 
           \end{array}  \right. \]
 Then $H(0,f) = f$ and $H(1,f) = \be^{-1}i_2\ga$, which is independent of $f$.

          

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