I know this fact from a homological algebra background, so forgive me if I am using a language that is foreign to you.

Easy Lemma: Let If $(X,d)$ be an exact chain complex of vector spaces over a field of characteristic $0$ $K$ and $e:X \to X$ a contraction, i.e. $de +ed =c $, with $0 \neq c \in $K$. Then $d+e :\bigoplus X_{2j} \to \bigoplus X_{2j-1}$ is an isomorphism.

If $e$ and $d$ are explicit, this is reasonably explicit. The appropriate chain complex for your problem is constructed as follows. Fix a vector space $V$ of finite dimension, write $E^p := \Lambda^{p} V^{\ast}$ and $S^q := Sym^q V^{\ast}$. Let $R^{\ast} := \bigoplus_{p,q} E^p \otimes S^q$. This is a graded commutative algebra if you give $E^p$ the degree $p$ and $S^q$ the degree $2q$.

There are canonical, mutually inverse, isomorphism $d:E^1 \to S^1$ and $e:S^1 \to E^1$. Extend $d$ to all of $R$ by the property $d(xy)= (dx)y + (-1)^{deg(x) } x (dy)$ and the requirement that $d(S^q)=0$. Do the same with $e$ (but here $e(E^p)=0$). The formulas $d^2=e^2=0$ hold.

Explicit formulas are

$$ d(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = \sum_{i=1}^{p} (-1)^{i-1} (v_1 \wedge v_{i-1} \wedge v_{i+1} \ldots v_p \otimes (dv_i w_1 \ldots w_q)) $$

and

$$ e(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = (-1)^p \sum_{i=1}^{q} (ew_i \wedge v_1 \wedge \ldots v_p \otimes ( w_1 \ldots w_{i-1} w_{i+1} \ldots w_q)). $$

Now on the piece $E^p \otimes S^q$, the formula $de+ed=p+q$ holds, as you prove using the product formulae and induction. The sequence (maps given by $d$)

$$ 0\to E^n \to E^{n-1} \otimes S ^1 \to \ldots E^1 \otimes S^{n-1} \to S^n \to 0. $$

is a chain complex and $e$ is a contraction, as in the above easy lemma. The sum $d+e$ is your desired isomorphism.

Remark: the construction is completely natural and therefore $GL(V)$-equivariant.

I know this fact from a homological algebra background, so forgive me if I am using a language that is foreign to you.

Easy Lemma: Let If $(X,d)$ be an exact chain complex of vector spaces over a field of characteristic $0$ $K$ and $e:X \to X$ a contraction, i.e. $de +ed =c $, with $0 \neq c \in K$. Then $d+e :\bigoplus X_{2j} \to \bigoplus X_{2j-1}$ is an isomorphism.

If $e$ and $d$ are explicit, this is reasonably explicit. The appropriate chain complex for your problem is constructed as follows. Fix a vector space $V$ of finite dimension, write $E^p := \Lambda^{p} V^{\ast}$ and $S^q := Sym^q V^{\ast}$. Let $R^{\ast} := \bigoplus_{p,q} E^p \otimes S^q$. This is a graded commutative algebra if you give $E^p$ the degree $p$ and $S^q$ the degree $2q$.

There are canonical, mutually inverse, isomorphism $d:E^1 \to S^1$ and $e:S^1 \to E^1$. Extend $d$ to all of $R$ by the property $d(xy)= (dx)y + (-1)^{deg(x) } x (dy)$ and the requirement that $d(S^q)=0$. Do the same with $e$ (but here $e(E^p)=0$). The formulas $d^2=e^2=0$ hold.

Explicit formulas are

$$ d(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = \sum_{i=1}^{p} (-1)^{i-1} (v_1 \wedge v_{i-1} \wedge v_{i+1} \ldots v_p \otimes (dv_i w_1 \ldots w_q)) $$

and

$$ e(v_1 \wedge \ldots v_p)\otimes (w_1 \ldots w_q) = (-1)^p \sum_{i=1}^{q} (ew_i \wedge v_1 \wedge \ldots v_p \otimes ( w_1 \ldots w_{i-1} w_{i+1} \ldots w_q)). $$

Now on the piece $E^p \otimes S^q$, the formula $de+ed=p+q$ holds, as you prove using the product formulae and induction. The sequence (maps given by $d$)

$$ 0\to E^n \to E^{n-1} \otimes S ^1 \to \ldots E^1 \otimes S^{n-1} \to S^n \to 0. $$

is a chain complex and $e$ is a contraction, as in the above easy lemma. The sum $d+e$ is your desired isomorphism.

Remark: the construction is completely natural and therefore $GL(V)$-equivariant.