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Orbits of the root action are precisely roots belonging to the same irreducible factor, proving transitivity iff irreducibility.

For a same-degree counterexample, let $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$,Both split in $L$, but the first action has two orbits of size two while the second is transitive, so the subgroups are not conjugate in $S_4$.

Over a finite field, Frobenius cycle type proves the assertion. The given quartic reduces over ${\mathbb{F}}_2$ to $X^4+X^3+1$, which is irreducible, so its group there is cyclic of order four generated by a $4$-cycle.

Over $\mathbb{Q}$ the polynomial is irreducible by the same reduction. Modulo $3$ it factors as linear times irreducible cubic, so Dedekind factorization cycle test gives a $3$-cycle; modulo $2$ it gives a $4$-cycle. Its discriminant is $229$, and modulo $23$ its factor degrees are $1,1,2$, giving a transposition. These cycle types force the transitive group to be $S_4$.

Put $\Gamma (z)=I(z)$ for $\Re z>0$. Integration by parts gives $\Gamma (z+1)=z\Gamma (z)$, both proving $\Gamma (n)=(n-1)!$ and defining the meromorphic continuation. A keyhole-contour evaluation of ${\int }_0^{\infty }{t}^{z-1}/(1+t)dt$ gives

Forthe reflection formula at $z$ and $z+\frac{1}{2}$ gives $g(z+1)-g(z)=-2\log 2$. Thus $h(z)=g(z)+2z\log 2$ is entire and $1$-periodic; poles and zeros have cancelled in the quotient. Its Fourier expansion and the exponential growth bound make it a finite Laurent polynomial in $e^{2\pi iz}$; the same growth bound for $e^h$ excludes every nonconstant mode. Hence $h=b$. At $z=1/2$, $e^b=2\sqrt{\pi }$, yielding the duplication formula.

Set $z=1/6$ and use reflection at $1/3$:Therefore