= Solution
Let $L$ be the <splitting field> of the <separable polynomial>. Its <Galois group> is $G=\operatorname{Aut}_K(L)$, acting faithfully on its $n$ distinct roots, hence as a subgroup of $S_n$. An automorphism preserves each root's minimal polynomial. Conversely an isomorphism $K(\alpha)\to K(\beta)$ for roots of the same irreducible factor extends to an embedding of $L$ in an <algebraic closure>, and normality of the <splitting field> makes the image $L$ itself. Thus the root orbits are precisely the irreducible factors' root sets. \b[The action is transitive exactly when the polynomial is irreducible.]
If $n=p$ is <prime>, a transitive action has $p\mid|G|$ by orbit-stabilizer, so <Cauchy theorem for groups> supplies an element of order $p$. Its permutation must be a $p$-cycle. Conversely a $p$-cycle already acts transitively on all $p$ roots. This proves the stated equivalence for <prime> degree.
The factorization theorem modulo a good <prime> says that if an <integer> polynomial retains its degree and has square-free reduction modulo $p$, and that reduction's irreducible factor degrees are $d_1,\ldots,d_r$, its <Galois group> over $\mathbb Q$ contains a permutation with disjoint cycle lengths $d_1,\ldots,d_r$. Equivalently the finite-field <Galois group>, generated by Frobenius on the roots, occurs as an unramified decomposition subgroup. The degree-retention qualification matters for nonmonic polynomials.
For the given quintic, <Eisenstein criterion> at three proves irreducibility. Modulo two it factors as
$$
x^5+x+1=(x^2+x+1)(x^3+x^2+1).
$$
Neither factor has a root in $\mathbb F_2$, so both are irreducible, and they are distinct; the reduction is separable. Thus $G$ contains an element of cycle type $(2)(3)$, whose cube is a <transposition>. It is also transitive of degree five.
To prove that this forces the full <symmetric group> here, form the <graph> on the five roots with an <edge> for each conjugate in $G$ of that <transposition>. The <graph> is $G$-invariant, and transitivity makes its connected components all the same size. That size divides five; there is an <edge>, so the size cannot be one. Hence the <graph> is connected. <Transpositions> along the <edges> of a connected <graph> generate all <transpositions> (move along a path and conjugate successive swaps), and hence generate $S_5$. They all belong to $G$, giving
$$
\boxed{G=S_5.}
$$
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