$\mathbb{Z}/2$-action on spectra given by inversion

This is a somewhat naive question to which I don't know the answer. There is a map of spectra $S \to S$, defined up to homotopy, given by multiplication by $-1$, and it satisfies the relation $(-1)^2 \simeq 1$. Can this be refined to a strict $\mathbb{Z}/2$-action on the sphere $S$ (and thus on all spectra)?

My feeling is that the answer is probably not, but I haven't been able to prove it. I do know that when one inverts $2$, there is a $\mathbb{Z}/2$-action: if I understand correctly, the obstructions to rigidifying a homotopy $\mathbb{Z}/2$-action to an actual $\mathbb{Z}/2$-action on a spectrum $X$ live in the $\mathbb{Z}/2$-cohomology groups of $B Aut(X)$ (where $Aut(X)$ is the monoid of homotopy self-equivalences of $X$), which should be zero when $2$ acts invertibly on $\pi_*X$.

-

There's often no self-map of $S$ which induces multiplication by $(-1)$, because it's usually not cofibrant-fibrant. There's no homotopical obstruction to it existing, because the map $E \mapsto E \wedge S^1$ is an autoequivalence of the stable homotopy category, and $\mathbb{Z}/2$ acts on $S^1$. Or, if you prefer, we a map of based $\mathbb{Z}/2$-spaces $$\mathbb{Z}/2_+ \to S^0$$ and taking the homotopy fiber of the induced map of suspension spectra, we get a homotopy fiber sequence of spectra with $\mathbb{Z}/2$-action $$S \to S \wedge \mathbb{Z}/2_+ \to S$$ with the first term having a negation action.
One can see visibly that in, e.g., the category of symmetric spectra, the sphere has no $-1$ self-map, but the equivalent free spectrum $F_1(S^1)$ does (for appropriate definitions of "S^1"). In the framework of Adams' "Stable homotopy and generalized homology", because maps are only defined "eventually", the analogue of $F_1(S^1)$ and $S$ are isomorphic objects, so you do have this map.
So if you want it on-the-nose, it depends on your framework. If you want it homotopically, it always exists (roughly because the map $BGL_1(S) \to B\mathbb{Z}/2$ splits).
Fair enough. You're absolutely right that there is a $\mathbb{Z}/2$-action on the sphere $S^n$ (considered as a topological space), and I'm embarrassed that I convinced myself that this wasn't true. Thanks! –  Akhil Mathew Sep 29 '12 at 21:09
And, to be fair, this is very close to facts which are false. There is not a $\mathbb{Z}/2$-action by $(-1)$ on the sphere which is really "central" in the sense that you can coherently move it across $\wedge$ as you like. –  Tyler Lawson Sep 29 '12 at 21:55
That's true. Somehow I imagined that having a $\mathbb{Z}/2$-action would make the stable category more "additive" than it is (before passage to homotopy), but I never thought about it on spaces. –  Akhil Mathew Oct 2 '12 at 12:00