Consider the finite field $\mathbb F_p$ and ring of Witt vectors $W(\mathbb F_p) \cong \mathbb Z_p$, where $\mathbb Z_p$ is the ring of $p$-adic numbers.
Why the Frobenius map on $W(\mathbb F_p)$ fix the prime $p$ ?
Let $a=(a_0,a_1,a_2, \cdots) \in W(\mathbb F_p)$, then the Frobenius $\varphi: W(\mathbb F_p) \to W(\mathbb F_p)$ is defined by $$\varphi(a)=(a_0^p,a_1^p, a_2^p, \cdots),~a_i \in \mathbb F_p.$$ But since $a_i^p=a_i$ in $\mathbb F_p$, we have $\varphi(a)=(a_0,a_1, \cdots)$, i.e., $\varphi$ is identity on $W(\mathbb F_p)$. Thus $\varphi$ fix $p$.
I think we can replace $\mathbb F_p$ by any characteristic $p$ finite field.
The same Frobenius endomorphism extends to the corresponding power series ring $W(\mathbb F_{p^n})[[x]]$ by $\varphi(x)=x^p$ i.e., $$\varphi (f(x))=f(x^p),~~f(x) \in W(\mathbb F_{p^n})[[x]],$$ where $\varphi$ is identity on $W(\mathbb F_p)$ but not identity on $W(\mathbb F_{p^n})$.
I appreciate your comments.
Maybe first of all it should be noted that if $R$ is any ring which contains the integers $\mathbb Z$, then any ring endomorphism of $R$, by virtue of fixing $1$, fixes all elements of $\mathbb Z$.
In this concrete case, one can also note that if $k$ is any perfect field of characteristic $p$, the element $p$ (i.e. $1+ \dots +1$, $p$ times) in $W(k)$ is actually
$$(0,1,0,0, \dots)$$
which quite obviously is fixed by the Frobenius map
$$(x_0, x_1, \dots ) \mapsto (x_0^p, x_1^p, \dots ) $$
regardless of what other elements might or might not be fixed by the Frobenius. In fact, it follows immediately that $\mathbb Z_p \simeq W(\mathbb F_p) \subset W(k)$ is exactly the fixed point set of the Frobenius map, because $\mathbb F_p \subset k$ is exactly the fixed point set under the classical Frobenius (raising to the $p$-th power).