Interesting things happen if you reduce this dynamical system modulo some integer $N$.

This makes sense because your map $s$ is compatible with reduction modulo $N$, and your map $e$ is $\mathbf{Z}$-linear, so $f$ induces a map $f_N : P_N \to P_N$ with $P_N = P \otimes \mathbf{Z}/N\mathbf{Z}$.

In the case $N=2$, what is the map $f_2$? Note that $P_2 = \mathbf{F}_2[x_1,\ldots,x_n]/(x_1^2-x_1,\ldots,x_n^2-x_n)$ can be identified with the ring of functions $\mathbf{F}_2^n \to \mathbf{F}_2$ (every such function is polynomial!). This ring is in bijection with the powerset of $\mathbf{F}_2^n$ : we can map any subset of $\mathbf{F}_2^n$ to its characteristic function. Now, the map $s_2: P_2 \to P_2$ has a very natural interpretation : it maps a subset $A$ of $\mathbf{F}_2^n$ to the subset $\cap_{\sigma \in \Sigma(n)} \sigma(A)$. Also, by definition, the image of $e_2 = e \otimes \mathbf{F}_2$ is contained in the set of affine hyperplanes of $\mathbf{F}_2^n$ (together with $\emptyset$ and $\mathbf{F}_2^n$).

So we want to determine which affine hyperplanes are fixed by $f_2$. This is not obvious at first sight, since the map $e_2$ doesn't seem to have a nice interpretation like $s_2$. But luckily, some linear algebra over $\mathbf{F}_2$ shows that there are only $6$ possibilities for $s_2(H)$ where $H$ is an affine hyperplane of $\mathbf{F}_2^n$. Namely $s_2(H)=H$ when $H$ is symmetric, and $s_2(H)$ can only be $\emptyset$, $\{0\}$, $\{1\}$ or $\{0,1\}$ otherwise. Here $0$ resp. $1$ denotes the constant vector $(0,\ldots,0)$ resp. $(1,\ldots,1)$. It's possible to compute the image by $e_2$ of these subsets, and after some computations we find that the only fixed points of $f_2$ are the polynomials $0$, $1$, $x_n$, together with $1+x_1+\cdots+x_{n-1}$ when $n$ is odd.

Now let's go back to $\mathbf{Z}$. Obviously every fixed point $p$ of $f$ must reduce to a fixed point of $f_2$ mod $2$. Conversely, we have a kind of Hensel lemma : every fixed point of $f_2$ lifts to a unique fixed point of $f_{\mathbf{Z}_2} : P \otimes \mathbf{Z}_2 \to P \otimes \mathbf{Z}_2$. Here $f_{\mathbf{Z}_2}$ is defined by just replacing the base ring $\mathbf{Z}$ by $\mathbf{Z}_2$ in the definition of $f$.

This lifting property is a consequence of the following fact : $f$ is contracting for the $2$-adic topology on every ball $p_0+2P$ with $p_0 \in P$. Indeed, assume $p=p_0+2q$ with $q \in P$. Then

\begin{equation*}
s(p) = \prod_{\sigma \in \Sigma(n)} p_0^\sigma + 2q^\sigma \equiv s(p_0)+2\sum_{\sigma} q^\sigma \cdot \prod_{\tau \neq \sigma} p_0^\tau \pmod{4}
\end{equation*}

Now, observe that the stabilizer $H$ of $\overline{p_0} \in P_2$ in $\Sigma(n)$ is isomorphic to $\Sigma(m) \times \Sigma(n-m)$ for some $0 \leq m \leq n$, so in particular $\operatorname{card}(H) \geq 2$. This implies that $\prod_{\tau \neq \sigma} \overline{p_0}^\tau = s(\overline{p_0})$ in $P_2$. Furthermore $\sum_{\sigma} \overline{q}^\sigma = 0$ since the stabilizer of $\overline{q} \in P_2$ has even order. It follows that $s(p) \equiv s(p_0) \pmod{4}$. The same argument shows that $p \equiv p_0 \pmod{2^m}$ implies $s(p) \equiv s(p_0) \pmod{2^{m+1}}$, so $s$, and thus $f$, is contracting.

It follows immediately that the only fixed points of $f$ in the balls $2P$, $1+2P$ and $x_n+2P$ are $0$, $1$ and $x_n$ respectively. It remains to consider the fixed point $1+x_1+\cdots+x_{n-1}$ with $n \geq 3$ odd. The corresponding fixed point of $f_{\mathbf{Z}_2}$ is obtained as the limit of $f^m(1+x_1+\cdots+x_{n-1})$ in $P \otimes \mathbf{Z}_2$ when $m \to \infty$.

**EDIT** This question is still surprising me. I thought this fixed point would not belong to $P$, but it does. In fact $1-x_1+x_2 -\cdots +x_{n-1} \in P$ is a fixed point of $f$ for any odd $n \geq 3$. This is because of the following identity in $P$

\begin{equation*}
\prod_{\sigma \in \Sigma(n)} 1-x_{\sigma(1)}+x_{\sigma(2)}- \cdots +x_{\sigma(n-1)} = 1-S_1+S_2-\cdots+S_{n-1}
\end{equation*}
which is true because it's true for any choice of $x_1,\ldots,x_n \in \{0,1\} \subset \mathbf{Z}$ (both sides are $0$ except when $x_1=\cdots=x_n=0$).

To sum up, for $n \geq 2$ the only fixed points of $f$ are $0$, $1$, $x_n$, together with $1-x_1+x_2-\ldots+x_{n-1}$ for odd $n$.

By the same method, one might determine the periodic orbits of $f$. Since $f$ is contracting on each ball, each $f$-orbit reduces mod $2$ to a $f_2$-orbit with the same period. The only non-trivial orbit of $f_2$ is $\{1+x_1,1+x_1+\cdots+x_n\}$. Again, this non-trivial orbit lifts to a unique orbit in $P \otimes \mathbf{Z}_2$, but I don't know whether this orbit is in $P$. I guess the next step would be to determine all preperiodic points of $f$.

This method also suggests to study the analogous dynamical system on $P = \mathbf{Z}[x_1,\ldots,x_n]/(x_1^p-x_1,\ldots,x_n^p-x_n)$ for any prime $p$.

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