Proof.
An automorphism of (K, \alpha ) is a map \varphi : K \to K in D(\mathcal{O}) with \varphi \otimes _\mathcal {O} \text{id}_{\mathcal{O}_0} = \text{id}. This is the same thing as saying that
K \xrightarrow {\varphi - \text{id}} K \to K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0
is zero. We conclude the group of automorphisms is the cokernel of a map
\mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(K, K_0[-1]) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(K, K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I})
by the distinguished triangle
K \otimes _\mathcal {O}^\mathbf {L} \mathcal{I} \to K \to K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0 \to (K \otimes _\mathcal {O}^\mathbf {L} \mathcal{I})[1]
in D(\mathcal{O}) and Derived Categories, Lemma 13.4.2. To translate into the groups in the lemma use adjunction of the restriction functor D(\mathcal{O}_0) \to D(\mathcal{O}) and - \otimes _\mathcal {O} \mathcal{O}_0 : D(\mathcal{O}) \to D(\mathcal{O}_0). This proves (1).
Proof of (2). Assume that K_0 = K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0 in D(\mathcal{O}). By Lemma 91.16.6 the map sending a lift (K', \alpha _0) to the obstruction o(\alpha _0) to lifting \alpha _0 defines a canonical injective map from the set of isomomorphism classes of pairs to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_0}(K_0, K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I}). To finish the proof we show that it is surjective. Pick \xi : K_0 \to (K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I})[1] in the \mathop{\mathrm{Ext}}\nolimits ^1 of the lemma. Choose a bounded above complex \mathcal{F}^\bullet of flat \mathcal{O}-modules representing K. The map \xi can be represented as t \circ s^{-1} where s : \mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet is a quasi-isomorphism and t : \mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}[1] is a map of complexes. By Lemma 91.16.5 we can assume there exists a quasi-isomorphism \mathcal{G}^\bullet \to \mathcal{F}^\bullet of complexes of \mathcal{O}-modules such that \mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet factors through s up to homotopy. We may and do replace \mathcal{G}^\bullet by a bounded above complex of flat \mathcal{O}-modules (by picking a qis from such to \mathcal{G}^\bullet and replacing). Then we see that \xi is represented by a map of complexes t : \mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}[1] and the quasi-isomorphism \mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet . Set
\mathcal{H}^ n = \mathcal{F}^ n \times _{\mathcal{F}_0^ n} \mathcal{G}_0^ n
with differentials
\mathcal{H}^ n \to \mathcal{H}^{n + 1},\quad (f^ n, g_0^ n) \mapsto (d(f^ n) + t(g_0^ n), d(g_0^ n))
This makes sense as \mathcal{F}_0^{n + 1} \otimes _{\mathcal{O}_0} \mathcal{I} = \mathcal{F}^{n + 1} \otimes _\mathcal {O} \mathcal{I} = \mathcal{I}\mathcal{F}^{n + 1} \subset \mathcal{F}^{n + 1}. We omit the computation that shows that \mathcal{H}^\bullet is a complex of \mathcal{O}-modules. By construction there is a short exact sequence
0 \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I} \to \mathcal{H}^\bullet \to \mathcal{G}_0^\bullet \to 0
of complexes of \mathcal{O}-modules. Exactly as in the proof of Lemma 91.16.4 one shows that this sequence induces an isomorphism \alpha _0 : \mathcal{H}^\bullet \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0 \to \mathcal{G}_0^\bullet in D(\mathcal{O}_0). In other words, we have produced a pair (\mathcal{H}^\bullet , \alpha _0). We omit the verification that o(\alpha _0) = \xi ; hint: o(\alpha _0) can be computed explicitly in this case as we have maps \mathcal{H}^ n \to \mathcal{F}^ n (not compatible with differentials) lifting the components of \alpha _0. This finishes the proof.
\square
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