TITLE: An orbit of symmetric polynomials
QUESTION [15 upvotes]: Consider the ring of polynomials $R:=\mathbb{Z}[x_1,x_2,x_3]$. Define the operators $E, I:R\rightarrow R$ by $Ef(x_1,x_2,x_3)=f(x_1-1,x_2,x_3)$ and the identity $If=f$.
Let $\mathcal{L}:R\rightarrow R$ be the operator given by
$$\mathcal{L}f=[(x_1+x_2)(x_1+x_3)E-x_1^2I]f.$$
Let $1$ stand for the constant function $f(x_1,x_2,x_3)=1$ and $\mathcal{L}^2f=\mathcal{L}(\mathcal{L}f)$, etc.

CLAIM. Experiments suggest that $\mathcal{L}^n1$ is always a symmetric polynomial in $R$. Any proof?

EDIT. This has found a resolution (see Pietro Majer's answer).
For example, $\mathcal{L} 1=e_2$ and $\mathcal{L}^21=e_2^2-e_1e_2+e_3$ where $e_1=x_1+x_2+x_3, e_2=x_1x_2+x_1x_3+x_2x_3, e_3=x_1x_2x_3$ are the standard elementary symmetric polynomials.

QUESTIONS. (EDIT) These did not find a definitive answer (apart from Brendan McKay's evidence and argument).
(1) Are there other orbits of symmetric polynomials under $\mathcal{L}$?
(2) Are there other non-trivial operators with similar property over $R$?
(3) What about over rings of many more variables?

REPLY [7 votes]: Noting Pietro's finite check, I can report that all symmetric polynomials $p(x_1,x_2,x_3)$ up to degree 18 inclusive such that $\mathcal{L}p$ is symmetric are linear combinations of $1,\mathcal{L}1,\mathcal{L}^21,\ldots\,$.
This strongly suggests that there are no others.
An observation that might lead to an elementary proof is that, up to degree 18, all polynomials $p$ such that both $p$ and $\mathcal{L}p$ are symmetric are uniquely determined by the coefficients of the powers of $e_2$ (i.e. the terms in the representation in the base $\{e_1,e_2,e_3\}$ which have the form $c e_2^k$).