Homotopy orbits, spectra and infinite loop spaces

Question

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 $$ (\Omega^\infty X)_{hG} \to \Omega^\infty(X_{hG})\,? $$

Answer 1

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 $\pi_n (*)$.