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The following exercise from D. Robinson, A Course in the Theory of Groups confuses me:

Find two isomorphic extensions of $\mathbb Z_3$ by $\mathbb Z_3 \times \mathbb Z_3$ which are not equivalent.

For two group extensions $N, E, G$ and $\overline N, \overline E, \overline G$ with injections $\mu : N \to E, \overline\mu : \overline N \to \overline E$ and projections $\varepsilon : E \to G, \overline\varepsilon : \overline E \to \overline G$ a homomorphism of extensions is a triple of homomorphism $(\alpha, \beta, \gamma)$ from $N$ to $\overline N$, $E\to\overline E$ and $G \to \overline G$ such that the resulting diagram commutes, see this image from the book.

An equivalence between extensions $N, E, G$ and $N, \overline E, G$ is a triple of morphisms $(1_N, \beta, 1_G)$, an isomorphism between extension $N, E, G$ and $\overline N , \overline E, G$ is a triple of morphisms $(\alpha, \beta, 1)$ where $\alpha$ (and hence $\beta$) are isomorphisms. Note that in an isomorphism in general $N \ne \overline N$ , where they are equal for the notion of equivalence.

Coming back to the exercise, first as $\mathbb Z_3$ is a field, the only isomorphism from $\mathbb Z_3$ to $\mathbb Z_3$ is the identity map. So, as I see it, there is no way to give an isomorphism $\alpha \ne 1_{\mathbb Z_3}$, hence not possible to give an isomorphism which is not also an equivalence?

Or do I misunderstand something? Could you please clarify what this exercise asks for?

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There is only one ring automorphism of $\Bbb Z_3$, but the topic at hand is group extensions, and so group automorphisms, and there is a non-identity group automorphism of $\Bbb Z_3$ (switching the images of $1$ and $2$).

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