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 $$ \Hom_\mathcal{O}(K, K_0[-1]) \longrightarrow \Hom_\mathcal{O}(K, K_0 \otimes_{\mathcal{O}_0}^\mathbf{L} \mathcal{I}) $$ 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 \ref{derived-lemma-representable-homological}. 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). \medskip\noindent Proof of (2). Assume that $K_0 = K \otimes_\mathcal{O}^\mathbf{L} \mathcal{O}_0$ in $D(\mathcal{O})$. By Lemma \ref{lemma-inf-obs-map-defo-complex} 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 $\Ext^1_{\mathcal{O}_0}(K_0, K_0 \otimes_{\mathcal{O}_0}^\mathbf{L} \mathcal{I})$. 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 $\Ext^1$ 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 \ref{lemma-lift-map-complexes} 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 \ref{lemma-canonical-class-obstruction} 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 explitly 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.