Let $p$ be a prime. The group $\operatorname{Aut}(\Bbb Z/p\Bbb Z)$ acts regularly on the set of generators of $\Bbb Z/p\Bbb Z$. Therefore, for every $i,j=1,\dots,p-1$, there is one and only one $\sigma_{ij}\in S_{p-1}$ such that: \begin{equation} \begin{cases} \sigma_{ij}(i)=j\space\space\text{ (by the regularity of the action)}\\ \sigma_{ij}(1)=ji^{-1}\space\space\text{ (by the linearity of }\sigma_{ij}) \\ \sigma_{ij}(k)\ne k,\text{ for every }k\space(i\ne j)\\ \Sigma=\{\sigma_{ij}, i,j =1,\dots,p-1\} \text{ is an abelian subgroup of order }p-1 \text{ of }S_{p-1} \text{ (so there are repetitions)} \tag1 \end{cases} \end{equation}
For example, for $p=5$ the normal Klein four-subgroup $\Sigma=\{(),(12)(34),(13)(24),(14)(23)\}\le S_4$, with the identifications: \begin{alignat}{1} &\sigma_{ii}=(),\text{ for }i=1,2,3,4 \\ &\sigma_{12}=\sigma_{21}=\sigma_{34}=\sigma_{43}=(12)(34) \\ &\sigma_{13}=\sigma_{31}=\sigma_{24}=\sigma_{42}=(13)(24) \\ &\sigma_{14}=\sigma_{41}=\sigma_{23}=\sigma_{32}=(14)(23) \\ \end{alignat} fulfils the first and third equations in $(1)$, but not the second (for example, $\sigma_{34}(1)=2$, but $4.3^{-1}=4.3^3=4.27=108=3$).
Besides, some elements of the non-normal Klein four-subgroups of $S_4$ fail to move all the generators of $\Bbb Z/5\Bbb Z$, so they do not fulfil $(1)$ either.
On the other hand, $\Sigma=\langle(1243)\rangle=$ $\{(),(1243),(14)(23),(1342)\}$ with the identifications: \begin{alignat}{1} &\sigma_{ii}=(),\text{ for }i=1,2,3,4 \\ &\sigma_{12}=\sigma_{24}=\sigma_{43}=\sigma_{31}=(1243) \\ &\sigma_{14}=\sigma_{41}=\sigma_{23}=\sigma_{32}=(14)(23) \\ &\sigma_{13}=\sigma_{34}=\sigma_{42}=\sigma_{21}=(1342) \\ \end{alignat} fulfils $(1)$. In fact: \begin{alignat}{1} &\sigma_{12}(1)=2\text{ and }2.1^{-1}=2.1=2 \\ &\sigma_{24}(1)=2\text{ and }4.2^{-1}=4.2^3=4.8=32=2 \\ &\sigma_{43}(1)=2\text{ and }3.4^{-1}=3.4=12=2 \\ &\sigma_{31}(1)=2\text{ and }1.3^{-1}=1.3^3=27=2 \\ & \\ &\sigma_{14}(1)=4\text{ and }4.1^{-1}=4.1=4 \\ &\sigma_{41}(1)=4\text{ and }1.4^{-1}=4 \\ &\sigma_{23}(1)=4\text{ and }3.2^{-1}=3.2^3=3.8=24=4 \\ &\sigma_{32}(1)=4\text{ and }2.3^{-1}=2.3^3=2.27=54=4 \\ & \\ &\sigma_{13}(1)=3\text{ and }3.1^{-1}=3 \\ &\sigma_{34}(1)=3\text{ and }4.3^{-1}=4.3^3=4.27=108=3 \\ &\sigma_{42}(1)=3\text{ and }2.4^{-1}=2.4=8=3 \\ &\sigma_{21}(1)=3\text{ and }1.2^{-1}=2^3=8=3 \\ \end{alignat} Therefore, $\operatorname{Aut}(\Bbb Z/5\Bbb Z)=\{\psi_i,i=1,2,3,4\}$, where $\psi_i(0)=0$ and $\psi_i(j)=(1243)^i(j)$ for $j=1,2,3,4$, and hence $\operatorname{Aut}(\Bbb Z/5\Bbb Z)\cong\Bbb Z/4\Bbb Z$.
My question: is $(1)$ constraining enough to force by itself $\Sigma$ to be cyclic (and hence $\operatorname{Aut}(\Bbb Z/p\Bbb Z)\cong\Bbb Z/(p-1)\Bbb Z$) for every $p$, like it does for $p=5$?