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I am asking for help on the following exercise:

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

as taken from D. Robinson, A Course in the Theory of Groups, see my other recent post for the notation.

There are five groups of order $27$. I found example extensions with the same groups that are not equivalent, but I was unable to give an isomorphism of extensions. Namely $E = \mathbb Z_3 \oplus \mathbb Z_9$ with $\mu(1) = (0,3)$ and $\overline \mu(1) = (1,3)$. Then an automorphism $\beta : E \to E$ giving equivalent extensions has to fulfill $\beta(0,3) = (1,3)$, but as $\beta(0,3) = \beta(0,1) + \beta(0,1) + \beta(0,1) = (0, x) \ne (1,3)$ this is not possible. The same argument works for the semidirect product $\mathbb Z_9 \rtimes \mathbb Z_3$, but both could not be made isomorphic extensions if we use the only non-trivial automorphism $\alpha(1) = 2$ on $\mathbb Z_3$. In $\mathbb Z_{27}$ we have only one way to embed $\mathbb Z_3$, so this will not work either. And $G = \mathbb Z_3 \oplus \mathbb Z_3 \oplus \mathbb Z_3$ will not work either, as we can alway write $\mu(Z_3) \oplus U = G$ as this is an $\mathbb Z_3$-vector space, hence make every isomorphism $\overline \mu(\mathbb Z_3) = \beta(\mu(\mathbb Z_3)$ into an automorphism of $G$. So the only promising candidate seems to be $UT(3,3)$, but here as every element has order three I am somewhat unable to go along the same lines above to show that certain maps could not be extended to isomorphisms.

So I am asking for help on this exercise! Any hints?

EDIT: It is not a duplicate question, the other had a very specific question concerning understanding what is asked for. This one asks for help on the exercise themselve. Should not be difficult to tell that they are both different!

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