TITLE: $RO(G)$-graded homotopy groups vs. Mackey functors
QUESTION [15 upvotes]: Everything here is model-independent: either take co/fibrant replacements wherever appropriate, or work $\infty$-categorically.
Also, I've looked through other similar MO questions, but I didn't find answers to the questions below.  Of course, please let me know if I missed anything!

In motivic homotopy theory, the category of motivic spectra comes with a set of "bigraded spheres", $S^{i,j} = \Sigma^i (\mathbb{G}_m)^{\wedge j}$.  It is well-known that homotopy classes of maps out of these don't detect equivalences.  This gives rise to a notion of "cellular" motivic spectra (as studied by Dugger--Isaksen): these sit as a right localization -- that is, a full subcategory whose inclusion admits a right adjoint -- which is by definition the largest subcategory in which bigraded homotopy groups do detect equivalences.
Back in equivariant homotopy theory, the analogous "tautological Picard elements" are the virtual representation spheres $S^V$, i.e. $V$ is a formal difference of finite-dimensional $G$-representations.  These corepresent what are therefore known as "$RO(G)$-graded homotopy groups".

A. If a map $X \to Y$ of $G$-spectra induces an isomorphism $[S^V,X] \xrightarrow{\cong} [S^V,Y]$ for all $V \in RO(G)$, is it necessarily an equivalence?

I would of course be interested to hear partial answers, e.g. if this is only known when $G$ is finite/discrete but is open for $G$ a (compact) Lie group -- similarly if this depends on what $G$-universe I'm working over (though the answer is pretty obviously "no" e.g. for the trivial universe).
Moreover, if the answer to the above is "no", then I have a follow-up question.

A'. What is known about the subcategory of cellular $G$-spectra?  Is it known to contain or not contain any particular $G$-spectra of interest?

Another question I have is the following.

B. Are there any particular $G$-spectra of interest with known $RO(G)$-graded homotopy groups?

In my limited understanding, the more common notion of "homotopy groups" in the equivariant world are given by homotopy classes of maps out of the stabilized orbits $\Sigma^n \Sigma^\infty_+ (G/H)$ for $n \in \mathbb{Z}$, which altogether assemble into a Mackey functor.  By definition, these detect equivalences.  (The motivic analog would be mapping out of $\Sigma^{\infty+n} X_+$ for all $X$ in the Nisnevich site.)

C. Besides the obvious ones, are there any known relationships between $RO(G)$-graded homotopy groups and the "homotopy" Mackey functor?

REPLY [4 votes]: A.  The brackets are the same computed in any model, as you say, and for most that entails 
fibrant approximation.   For genuine $G$-spectra (complete universe), $G$ a compact 
Lie group, it goes back to LMS http://www.math.uchicago.edu/~may/BOOKS/equi.pdf
that a map is a weak $G$-equivalence if it induces an isomorphism on $Z$--graded homotopy
groups, so on the $\pi_n^H$ for all integers $n$ and (closed) subgroups $H$.   That is the right 
definition of weak $G$-equivalence by Theorem I.4.6 there, which says that a map is a weak
$G$-equivalence if and only if it is a spacewise weak $G$-equivalence.   Taking $H=G$, one
is looking at trivial representations, and so the question amounts to asking when looking at all spheres $S^V$ for $V\in RO(G)$  is equivalent to looking at all $G/H_+\wedge S^n$.  Justin's
answer says that this is true for $G = Z/2$ and false for $G=Z/p$ for an odd prime $p$.
A'.  For G-CW spectra, as defined in LMS where all $G$-spectra are fibrant, a map is a weak 
$G$-equivalence if and only if it is a $G$-homotopy equivalence. For other models, one must 
first fibrant approximate and then construct $G$-CW spectra, giving the right model independent
notion.  Such $G$-CW spectra are not generally well related to model theoretic cell $G$-spectra.      See Chapter 24 of 
http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf
for discussion.
B.  Very few, as Justin says.  Don't have time to think this through right now.
C.  Question is cryptic.  Spell it out and I can try to answer.  Not clear what you have in mind.