Let $X$ be an (naive) $O(n)$-spectrum (I'm choosing to work with orthogonal spectra). I've recently come across the following results,


$$(S^{n-1} \wedge X)\_{hO(n)} \simeq X\_{hO(n-1)}$$


and


$$\Omega^\infty (X\_{hG)}) \simeq (\Omega^\infty X)\_{hG} $$


where $G$ is a compact Lie group.


I've spent the last few days trying to find sources for these results - and trying to prove them myself to little avail. 


Any help/suggestions or references would be greatly appreciated.




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**Added** The answer by Tyler has shown that the second is false. This raises the question about how close can we get, in the following sense:
If $X$ is $n$-connected, then how connected is the map
$$ 
(\Omega^\infty X)\_{hG} \to \Omega^\infty(X\_{hG})\,?
$$



Homotopy orbits, spectra and infinite loop spaces
Both of these are false.


The first is close to true: if $S(n-1)$ is the unit sphere in $\Bbb R^{n}$ with its standard $O(n)$-action, then we can identify $S(n-1)$ with $O(n) / O(n-1)$ and so get the identification
$$
(S(n-1)\_+ \wedge X)\_{hO(n)} = EO(n)\_+ \wedge\_{O(n)} O(n)/O(n-1)\_+ \wedge X = EO(n)\_+ \wedge\_{O(n-1)} X.
$$




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To show that the first one is false, let's take $X = \Sigma^\infty O(n)\_+$. Then the first equation would give an equivalence between $\Sigma^\infty S^{n-1}$ and $\Sigma^\infty O(n)/O(n-1)\_+ \cong \Sigma^\infty S(n-1)\_+$. These have nonisomorphic homology for all $n$.


For the second one, we can take $X$ to be the desuspension of the Eilenberg-MacLane spectrum $H\Bbb Z$ with trivial $O(n)$-action. Then $\Omega^\infty X$ is contractible and so its homotopy orbit space is $BG$ (or contractible, if you use the based orbit space), while
$$
\begin{align\*}
\pi\_n \Omega^\infty (\Sigma^{-1} H\Bbb Z)\_{hG}
&= \pi\_n (\Sigma^{-1} H\Bbb Z\_{hG})\\
&= \pi\_{n+1} (H\Bbb Z\_{hG})\\
&= \pi\_{n+1} (H\Bbb Z \wedge \Sigma^\infty BG\_+)\\
&= H\_{n+1}(BG, \Bbb Z)
\end{align\*}
$$
which is typically very different from either $\pi\_n BG$ or $\pi\_n (\*)$.


