ScalingStacks

Proof. We observe that the existence of an equivalence $\tau_{\leq 0}\mathcal{C}\simeq\gaunt_{n}$, Lemma 4.5, Lemma 4.10, and Corollary 4.14 guarantee that $\Aut((\tau_{\leq 0}\mathcal{C})^{\omega})$ in $\Fun((\tau_{\leq 0}\mathcal{C})^{\omega},(\tau_{\leq 0}\mathcal{C})^{\omega})$ is equivalent to the discrete group $(\mathbb{Z}/2)^{\times n}$. It therefore suffices to exhibit an equivalence of $\infty$-categories $\Aut(\mathcal{C})\simeq\Aut((\tau_{\leq 0}\mathcal{C})^{\omega})$.

Clearly $\Aut(\mathcal{C})$ is contained in the full subcategory $\Fun^{\mathrm{L}}(\mathcal{C},\mathcal{C})\subset\Fun(\mathcal{C},\mathcal{C})$ spanned by those functors that preserve small colimits. Since $(\tau_{\leq 0}C)^{\omega}$ is dense in $\mathcal{C}$, it follows from Lemma 9.2 that the inclusion $(\tau_{\leq 0}C)^{\omega}\hookrightarrow C$ induces a fully faithful functor

$$\Fun^{\mathrm{L}}(\mathcal{C},\mathcal{C})\hookrightarrow\Fun((\tau_{\leq 0}\mathcal{C})^{\omega},\mathcal{C}).$$

Moreover, any autoequivalence of $\mathcal{C}$ restricts to an autoequivalence of $\tau_{\leq 0}\mathcal{C}$ and hence an autoequivalence of $(\tau_{\leq 0}\mathcal{C})^{\omega}$. Thus restriction furnishes us with a fully faithful functor from $\Aut(\mathcal{C})$ to $\Aut((\tau_{\leq 0}\mathcal{C})^{\omega})\simeq(\mathbb{Z}/2)^{n}$.

It remains to show that the restriction functor is essentially surjective. For this, suppose $(\tau_{\leq 0}\mathcal{C})^{\omega}\to(\tau_{\leq 0}\mathcal{C})^{\omega}$ an autoequivalence. One may form the left Kan extension $\Phi\colon\mathcal{C}\to\mathcal{C}$ of the composite

$$\phi\colon(\tau_{\leq 0}\mathcal{C})^{\omega}\to(\tau_{\leq 0}\mathcal{C})^{\omega}\hookrightarrow\mathcal{C}$$

along the inclusion $(\tau_{\leq 0}\mathcal{C})^{\omega}\hookrightarrow\mathcal{C}$. One sees immediately that $\Phi$ is an equivalence, and moreover its restriction to $(\tau_{\leq 0}\mathcal{C})^{\omega}$ coincides with $\phi$. ∎