 Solution

Over ${\mathbb{F}}_3$,Both quadratics have discriminant $2$, a nonsquare in ${\mathbb{F}}_3$, so are irreducible polynomials. Both split in the same quadratic finite field ${\mathbb{F}}_9$. Consequently $\boxed{\mathrm{Gal}(t^4+1/{\mathbb{F}}_3)\cong C_2}$. Its generator is the finite-field Frobenius automorphism $x\mapsto x^3$, acting as a product of two transpositions on the four roots.