Assume $C\subset \mathbb{P}^2$ is a smooth cubic curve. Then there is a cyclic triple cover $\pi: S\rightarrow \mathbb{P}^2$ ramified in $C$. Let $\sigma$ be the covering automorphism (sheet interchange automorphism) associated to $\pi$, i.e. $S/<\sigma>\cong\mathbb{P}^2$, here $\sigma^3=id$. If $H:=\pi^{\*}l$ denotes the pullback of a line, then the canonical divisor of $S$ is $K\_S=-H$ and $K\_S^2=3$. So $S$ can also be seen as the blow up of $\mathbb{P}^2$ in 6 points and is therefore a cubic surface with the famous 27 lines on it.


Looking at the triple cover contruction the lines can be found the following way: $C$ has 9 points of inflection. The preimage of tangent line at such a point decomposes as $\pi^{-1}(l)=E\cup\sigma(E)\cup\sigma^2(E)$, where these are 3 (-1)-curves on $S$, so we get 9$\times$3=27 lines on $S$.


If $\pi: S\rightarrow \mathbb{P}^2$ is the triple cover and we pick 6 mutually skew lines $E\_1,\cdots,E\_6$ in the preimages of inflection lines, then there is a map $\phi: S \rightarrow \mathbb{P}^2$ such that $S$ is the blow up of $\mathbb{P}^2$ in 6 points $P\_1,\cdots,P\_6$ and the $E\_i$ are the exceptional curves. The strict transforms of the lines in $\mathbb{P}^2$ containing two different points $P\_i$ and $P\_j$, $1\leq i < j \leq 6$ give 15 (-1)-curves $F\_{i,j}$ on $S$. Finally there are six strict transforms of the conics $G\_i$ in $\mathbb{P}^2$ containing the $P\_j$ for $j\neq i$, $1\leq i \leq 6$.


What can we say about the images of the 27 lines under the automorphism $\sigma$? For example if we pick $E\_1$ can we say which lines $\sigma(E\_1)$ and $\sigma^2(E\_1)$ are in terms of the $F\_{i,j}$ and $G\_j$, e.g something like $\sigma(E\_1)=G\_1$? Or is there any other description which tells us exactly which 3 lines are in a preimage of an inflection line?


Background: I recently learned about the "Geiser involution": if one has a double cover $Y$ of $\mathbb{P}^2$ ramified in a smooth quartic $Q$, then $Y$ is the blow up of $\mathbb{P}^2$ in 7 points. $Y$ has 56 (-1)-curves, which arise in the preimages of the 28 bitangents to $Q$. The covering automorphism associated to $Y$ is called the "Geiser involution" and one can describe the images of the 56 (-1)-curves under this involution, see for example arxiv.org/pdf/math/0403245.pdf, Remark 3.3. So i was wondering if there is such a description for a cubic surface and its 27 lines.



How to find the action of an automorphism on the 27 lines on a cubic surface?
It is true that the surface obtained is a special one (for example, a general cubic surface admits no automorphism). In fact, taking coordinates $X,Y,Z$ on $\mathbb{P}^2$, the cubic curve is given by $F(X,Y,Z)$ for some polynomial of degree $3$ and the equation of the surface is $W^3=F(X,Y,Z)$. The automorphism corresponds to send $(W:X:Y:Z)$ onto $(aW:X:Y:Z)$ where $a$ is a $3$-rd root of unity.


As you said, the orbit of a line $E\subset S$ of the surface consists of three lines $E, \sigma(E), \sigma^2(E)$ with $E+\sigma(E)+\sigma^2(E)=\pi^{-1}(l)$ where $l$ is an inflexion line of $\mathbb{P}^2$. 
Note that $E+\sigma(E)+\sigma^2(E)$ is equal to the trace of an hyperplane of $\mathbb{P}^3$, in particular each intersects transversally the two others: $E\cdot \sigma(E)=E\cdot \sigma^2(E)=\sigma(E)\cdot \sigma^2(E)=1$.


I will use the same notation as you and write $\phi\colon S\to \mathbb{P}^2$ a birational morphism which is the blow-up of $6$ points. I denote by $E\_1,\dots,E\_6$ the six curves contracted and by $L\in \mathrm{Pic}(S)$ the pull-back of a general line of $\mathbb{P}^2$. Then $\mathrm{Pic}(S)$ is isomorphic to $\mathbb{Z}^7$ with basis $L,E\_1,\dots,E\_6$. The $27$ lines correspond to: $E\_i$, $i=1,...,6$, $F\_{ij}=L-E\_i-E\_j$ (line through two points) for $i\not=j$ and $G\_j=2L-\sum\_{i\not= j} E\_i$ (conics through $5$ points).


Because $\sigma(E\_1)\cdot E\_1=\sigma^2(E\_1)\cdot E\_1=1$, $\sigma(E\_1),\sigma^2(E\_1)$ are equal to $F\_{1i}$ or $G\_j$ for $i,j\not=1$. Because $\sigma^2(E\_1)\cdot \sigma(E\_1)=1$, we have moreover $i=j$. The orbit of $E\_1$ is thus $(E\_1,F\_{1i},G\_i)$ for some $i\not=1$.


Doing the same for $E\_2,...,E\_6$ we find a permutation $\tau$ of $(1,...,6)$ without fix points such that the orbit of $E\_i$ is {$E\_i, F\_{i,\tau(i)},G\_{\tau(i)}$} for $i=1,...,6$.


Because the image by $\sigma$ and $\sigma^2$ of the set {$E\_1,...,E\_6$} is $6$ skew lines, we can find that these two sets contain three of the $G\_i$ and three $F\_{i,j}$. Up to permutation of the curves $E\_i$, we have then
$E\_1\to F\_{12}\to G\_2$


$E\_2\to F\_{23}\to G\_3$


$E\_3\to F\_{13}\to G\_1$


$E\_4\to G\_5\to F\_{45}$


$E\_5\to G\_6\to F\_{56}$


$E\_6\to G\_4\to F\_{46}$


The image of the other lines is induced by these maps. We can for example easily write the matrix of $\sigma$ relatively to the basis $L,E\_1,...,E\_6$. (in particular, it corresponds to a birational map of $\mathbb{P}^2$ of degree $4$).


The matrix is 


$[4,1,1,1,2,2,2]$


$[-2,-1,0,-1,-1,-1,-1]$


$[-2,-1,-1,0,-1,-1,-1]$


$[-2,0,-1,-1,-1,-1,-1]$


$[-1,0,0,0,-1,-1,0]$


$[-1,0,0,0,0,-1,-1]$


$[-1,0,0,0,-1,0,-1]$


