ScalingStacks

Proof. First we observe that the $k$-cell $C_{k}\in\mathbb{G}_{n}$ is the unique object such that, up to isomorphism, there exists precisely $k$ other objects of $\mathbb{G}_{n}$ which occur as proper retracts (namely all the cells $C_{i}$ with $0\leq i<k$). Consequently, every autoequivalence $F$ of $\mathbb{G}_{n}$ must fix the objects.

Next we observe that for each $i<j$, there exists precisely one epimorphism $C_{j}\twoheadrightarrow C_{i}$. Since epimorphisms are preserved by any equivalence of categories, this unique epimorphism is also preserved by $F$. Similarly, for each $0\leq i<n$, there exist precisely two monomorphisms $C_{i}\hookrightarrow C_{n}$; these are either preserved by $F$ or else they are permuted. Thus every autoequivalence determines an element $\gamma(F)\in(\mathbb{Z}/2)^{n}$ such that $\gamma(F)_{i}=0$ just in case the pair of monomorphisms $C_{i}\hookrightarrow C_{n}$ is preserved by $F$, and $\gamma(F)_{i}=1$ just in case the pair of monomorphisms $C_{i}\hookrightarrow C_{n}$ is permuted by $F$. Note that of course $\gamma(\rho(g))=g$.

To conclude the proof, we observe that the symbol $\gamma(F)$ determines $F$. Indeed, every morphisms $C_{i}\to C_{j}$ in $\mathbb{G}_{n}$ admits a factorization

$$C_{i}\twoheadrightarrow C_{k}\hookrightarrow C_{n}\twoheadrightarrow C_{j}$$

for some $C_{k}\in\mathbb{G}_{n}$. ∎