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I was trying to find some actions that are permutation isomorphic, but not equivalent. See my recent post here for the definitions.

One natural candidate seems $A_4$. As the subgroups $U_1 = \langle (1,2,3)\rangle$ and $U_2 = \langle (1,2,4) \rangle$ are not conjugate in $A_4$, they induce non-equivalent actions on four points (their cosets). There is no normal subgroup in both subgroup, hence these actions are faithful and give two embeddings into $S_4$. As the alternating group is the only index two subgroup in $S_4$, these embeddings must both be equal, hence are isomorphic.

I computed them explicitly, set: $$ \alpha_1 := U_1, \quad \alpha_2 := U_1 (12)(34), \quad \alpha_3 := U_1 (13)(24), \quad \alpha_4 := U_1 (14)(23) $$ and $$ \beta_1 := U_2, \quad \beta_2 := U_2 (12)(34), \quad \beta_3 := U_2 (13)(24), \quad \beta_4 := U_2 (14)(23). $$ Then the permutations the elements of $A_4$ induce on these cosets are: $$ \begin{array}{l|l|l} (1 2 3) & (\alpha_2 \alpha_4 \alpha_3) & (\beta_1 \beta_3 \beta_4) \\ (1 3 2) & (\alpha_2 \alpha_3 \alpha_4) & (\beta_1 \beta_4 \beta_3) \\ (1 2 4) & (\alpha_1 \alpha_4 \alpha_3) & (\beta_2 \beta_3 \beta_4) \\ (1 4 2) & (\alpha_1 \alpha_3 \alpha_4) & (\beta_2 \beta_4 \beta_3) \\ (1 3 4) & (\alpha_1 \alpha_4 \alpha_2) & (\beta_1 \beta_2 \beta_3) \\ (1 4 3) & (\alpha_1 \alpha_2 \alpha_4) & (\beta_1 \beta_3 \beta_2) \\ (2 3 4) & (\alpha_1 \alpha_3 \alpha_2) & (\beta_1 \beta_2 \beta_4) \\ (2 4 3) & (\alpha_1 \alpha_2 \alpha_3) & (\beta_1 \beta_4 \beta_2) \\ (1 2)(34) & (\alpha_1 \alpha_2)(\alpha_3\alpha_4) & (\beta_1\beta_2)(\beta_3\beta_4) \\ (1 3)(24) & (\alpha_1 \alpha_3)(\alpha_2\alpha_4) & (\beta_1\beta_3)(\beta_2\beta_4) \\ (1 4)(23) & (\alpha_1 \alpha_4)(\alpha_2\alpha_3) & (\beta_1\beta_4)(\beta_2\beta_3) \end{array} $$ I worked out the identification $\psi(1) = \alpha_4, \psi(2) = \alpha_3, \psi(3) = \alpha_2, \psi(4) = \alpha_1$ (I just looked at the support of the $3$-cycles, for example $1 \in \{\alpha_2,\alpha_3,\alpha_4\}\cap\{\alpha_1,\alpha_3,\alpha_4\}\cap\{\alpha_1,\alpha_2,\alpha_4\}$ to find the images). Equivalently I found $$ \rho(1) = \beta_3, \rho(2) = \beta_4, \rho(3) = \beta_1, \rho(4) = \beta_2 $$ and similar $$ \sigma(\alpha_1) = \beta_2, \sigma(\alpha_2) = \beta_1, \sigma(\alpha_3) = \beta_4, \sigma(\alpha_4) = \beta_3 $$ and it turns out that under $\sigma$ both actions are equivalent!???

To see this, look at the following table, here I have written in the 2nd column some mapping in the $\alpha$'s, in the first are the corresponding elements of $A_4$ that realize this mapping of the $\alpha$'s, and in the 3rd column is what mapping should then be realised in the $\beta$'s. And this could be compared with the actual mapping in the above table, if they are equal, then we have $$ \sigma(\alpha^g) = \sigma(\alpha)^g $$ for $g \in A_4, \alpha \in \{\alpha_1, \ldots, \alpha_4\}$. $$ \begin{array}{lll} \mbox{elements from $A_4$} & \mbox{action on $U_1$; } \alpha_i \to \alpha_j & \sigma(\alpha_i) \to \sigma(\alpha_j) \\ (143), (12)(34) & \alpha_1 \to \alpha_2 & \beta_2 \to \beta_1 \\ (234), (13)(24) & \alpha_1 \to \alpha_3 & \beta_2 \to \beta_4 \\ (124), (14)(23) & \alpha_1 \to \alpha_4 & \beta_2 \to \beta_3 \\ (134), (12)(34) & \alpha_2 \to \alpha_1 & \beta_1 \to \beta_2 \\ (132), (14)(23) & \alpha_2 \to \alpha_3 & \beta_1 \to \beta_4 \\ (143), (13)(24) & \alpha_2 \to \alpha_4 & \beta_1 \to \beta_3 \\ (124), (13)(24) & \alpha_3 \to \alpha_1 & \beta_4 \to \beta_2 \\ (234), (14)(23) & \alpha_3 \to \alpha_2 & \beta_4 \to \beta_1 \\ (142), (12)(34) & \alpha_3 \to \alpha_4 & \beta_4 \to \beta_3 \\ (142), (14)(23) & \alpha_4 \to \alpha_1 & \beta_3 \to \beta_2 \\ (132), (13)(24) & \alpha_4 \to \alpha_2 & \beta_3 \to \beta_1 \\ (123), (12)(34) & \alpha_4 \to \alpha_3 & \beta_3 \to \beta_4 \end{array} $$ As I see it we have:

$$ \begin{array}{ccccc} \{\alpha_i\}&\rightarrow^g& \{\alpha_i\}\\ \downarrow_\sigma&&\downarrow_\sigma&&\\ \{\beta_i\}&\rightarrow^{g}&\{\beta_i\} \end{array} $$ so that the actions are equivalent. But the theory said they should not be equivalent. What went wrong here????

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    $\begingroup$ Are you sure that $U_1$ and $U_2$ are not conjugate in $A_4$? What's $(3,4)(1,2,3)(3,4)$? $\endgroup$ – David C. Ullrich Jan 13 '16 at 15:07
  • $\begingroup$ But $(3,4) \notin A_4$? $\endgroup$ – StefanH Jan 13 '16 at 15:08
  • $\begingroup$ Never mind - had the definition of $A_4$ backwards... sorry. $\endgroup$ – David C. Ullrich Jan 13 '16 at 15:10
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    $\begingroup$ $U_1$ and $U_2$ are Sylow $3$-subgroups of $A_4$ so they are certainly conjugate. $(1,2,3)$ is not conjugate to $(1,2,4)$, but it is conjugate to $(1,4,2)$. $\endgroup$ – Derek Holt Jan 13 '16 at 15:40
  • $\begingroup$ Oh thank you! Yes, of course... I was so confused and sticked to the thought that they are not conjugate. O_o $\endgroup$ – StefanH Jan 13 '16 at 15:49

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