How to construct an isomorphism $\wedge^2 V \xrightarrow{\sim} \wedge^2 V\otimes U^{\prime}$ as representations of $S_5$? Let $S_5$ be the symmetric group and $U^{\prime}$ be the $1$-dimensional alternating representation. Let $V$ be the $4$-dimensional standard representation, i.e.
$$
V=\{(x_1,x_2,x_3,x_4,x_5)\in \mathbb{C}^5\mid x_1+x_2+x_3+x_4+x_5=0\}
$$
and $S_5$ acts in the obvious way.
We consider the wedge product representation $\wedge^2 V$. By computing characters, we can show that
$\wedge^2 V\cong \wedge^2 V\otimes U^{\prime}$ as representations of $S_5$.

My question is: how to construct an isomorphism from $\wedge^2 V$ to $\wedge^2 V\otimes U^{\prime}$ explicitly?

 A: Question: "My question is: how to construct an isomorphism from ∧2V to ∧2V⊗U′ explicitly?"
Answer: It seems there is an isomorphism $\wedge^4 V \cong U'$
hence there is a canonical map
$$\phi: \wedge^2 V \otimes \wedge^2 V \rightarrow U'.$$
There are equalities
$$Hom(\wedge^2 V \otimes \wedge^2 V, U')\cong Hom(\wedge^2 V, Hom(\wedge^2 V,U')) \cong Hom(\wedge^2 V, (\wedge^2 V)^* \otimes U')$$
hence you get a canonical map
$$\phi^*: \wedge^2 V \rightarrow (\wedge^2 V)^* \otimes U'.$$
There is by Fulton/Harris an isomorphism $V^* \cong V$ and if $(\wedge^2 V)^* \cong \wedge^2(V^*) \cong \wedge^2 V$ you get a canonical map
$$ \psi: \wedge^2 V \rightarrow \wedge^2 V \otimes U'.$$
If you look in Fulton/Harris pages 27 and 476 you will find a character table for $S_5$ and an explicit isomorphism $(\wedge^2 V)^* \cong \wedge^2(V^*)$. I have not checked the details in FU, but if everything is correct you should get an explicit formula for the map $\psi$.
There is a map
$$\rho: \wedge^n(V^*) \rightarrow (\wedge^n V)^*$$
defined by
$$\rho(\phi_1 \wedge \cdots \wedge \phi_n)(v_1\wedge \cdots \wedge v_n):=det(\phi_j(v_i))$$
and on page 476 in FU this map is proved to be an isomorphism in characteristic zero.
