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 $E_1\to F_{12}\to G_2$

$E_2\to F_{23}\to G_3$

$E_3\to F_{13}\to G_1$

$E_4\to F_{45}\to G_5$

$E_5\to F_{56}\to G_6$

$E_6\to F_{46}\to G_4$

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 $4$).

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

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

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

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

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

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

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

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