FALSE: equivalent extensions mean only that the middle groups are isomorphic

The map $\pi$ is surjective, with kernel $\{0,\bar{3},\bar{6}\}$. Both $i_1$ and $i_2$ identify $C_3$ with that kernel, so $1\to C_3\stackrel{i_1}{\to }E\stackrel{\pi }{\to }{C}_3\to 1,1\to C_3\stackrel{i_2}{\to }E\stackrel{\pi }{\to }{C}_3\to 1$ are two extensions with the same middle group $E$.

Any automorphism of $E=C_9$ has the form ${\varphi }_u(\bar{x})=u\bar{x}$ with $u\in (\mathbb{Z}\mathrm{/}9\mathbb{Z}{)}^{\times }$. If an extension equivalence $\varphi$ existed, then [L1] would force $\varphi \circ i_1=i_2$ and $\pi \circ \varphi =\pi$. The first identity gives $3u\equiv 6(\mathrm{m}\mathrm{o}\mathrm{d}9)$, so $u\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)$. The second identity gives $u\bar{x}\equiv \bar{x}(\mathrm{m}\mathrm{o}\mathrm{d}3)$ for every $\bar{x}$, hence $u\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$. This is impossible. So no extension equivalence can exist.

By [L1], the two extensions from step 1.1 are therefore not equivalent even though they have the same middle group. The statement is false.