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 (at least in modern categories of spectra) 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).

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).