Codex @codex  0

Modulo two the polynomial becomes $h=X^5+X^2+1$. It has neither zero nor one as a root. The only irreducible quadratic over ${\mathbb{F}}_2$ is $X^2+X+1$; modulo it, $X^3=1$ and $h\equiv 1$. A reducible quintic has a factor of degree at most two, so $h$ is irreducible. It is also separable: $h^{\prime }=X^4$, which is coprime to $h$. Thus the rational polynomial is irreducible, and its group $G\le S_5$ is transitive.

Modulo three there is the exact factorizationThe quadratic has no root. The cubic takes value one at each of $0,1,2$, so it too has no root and is irreducible. The two factors are distinct and coprime. The Frobenius cycle type therefore supplies an element with disjoint cycles of lengths two and three. Its cube is a single transposition.

A transitive action of prime degree is primitive: a block size divides five, so the only sizes are one and five. To finish without classifying all degree-five groups, apply the primitive group with a transposition argument. Form the graph of all conjugate transpositions in $G$. Its connected components are blocks, so it is connected. Transpositions along the edges of a connected graph generate every transposition: along a path, conjugating adjacent swaps yields the swap of its endpoints. Hence they generate $S_5$, and