A natural identification of the second exterior power of linear operators ? Let $K$ be a field of characteristic zero and $V=K^n$. Let $A\in M(n,K)$, so we can think of $A$ as a linear map $A: V \to V$ be a linear map . Let $\wedge^2 V$ be the second exterior power of $V$ and $\wedge^2A: \wedge^2 V \to \wedge^2 V$ is the linear map defined as $\wedge^2A(x\wedge y)=A(x)\wedge A(y),\forall x,y\in V$ and extend it to whole of $\wedge ^2V$ linearly.
Now we can identify $\wedge^2V$ with the space of all $n\times n$ skew symmetric matrices with entries in $K$ (call this $Sk(n))$, with $e_i \wedge e_j$ (where $i<j$) is identified with that skew-symmetric matrix whose $ij$-th entry is $1$ and $ji$-th entry is $-1$ and all other entries are $0$. So the set of all such matrices makes a basis for the space of skew symmetric matrices . Denote this natural isomorphism $\wedge^2 V\to Sk(n)$ by $\phi$. 
Now consider a linear map $A^\wedge : Sk(n) \to Sk(n)$ as $A^\wedge (X)=AXA^t,\forall X \in Sk(n)$. 
My question is: Is $A^\wedge =\phi \circ \wedge^2A\circ \phi^{-1}$ ?
 A: What is really happening here is an (unnatural, since it requires a basis) identification of antisymmetric tensors in $V \otimes V$ with skew-symmetric matrices $M(n, k)$, and then just interpreting various natural maps in terms of these matrices. It is easiest to "undo" the identification with matrices and work naturally.
Let $X \in V \otimes V$ be any tensor. Once we choose a basis $(e_1, \ldots, e_n)$ for $V$, then $V \otimes V$ has a basis $\{e_i \otimes e_j \mid 1 \leq i, j \leq n \}$, and hence we also have a dual basis $\{e_i^* \otimes e_j^*\}$. We can record the tensor $X$ relative to this basis into a matrix $[X]$, where $[X]_{ij} := (e_i^* \otimes e_j^*)(X)$ is the coefficient of $e_i \otimes e_j$ in $X$. So we have an (unnatural) identification of the elements of $V \otimes V$ with $M(n, K)$, via the basis.
If $A: V \to V$ is a linear map, then $A$ extends to a linear map $A^{\otimes 2}$ on $V \otimes V$, defined by $A^{\otimes 2} (v \otimes w) = Av \otimes Aw$. We can check what happens to the matrix of $X$ under this map:
$$ \begin{aligned}\,
[A^{\otimes 2} X]_{ij}
&= (e_i^* \otimes e_j^*)\left(\sum_{l, k} [X]_{lk} A e_l \otimes A e_k\right) \\
&= \sum_{l, k} [X]_{lk} \cdot e_i^*(A e_l) \cdot e_j^*(A e_k) \\
&= \sum_{l, k} [A]_{il} [X]_{lk} [A]_{jk}
\end{aligned}$$
which gives that $[A^{\otimes 2} X] = [A][X][A]^t$, which looks exactly like your map $A^\vee$, albeit defined on all matrices, rather than skew-symmetric ones.
Next, we have a natural embedding $a: \wedge^2 V \to V \otimes V$, defined by $a(v \wedge w) = v \otimes w - w \otimes v$. You can check that the matrix $[a(e_i \wedge e_j)]$ is the skew-symmetric matrix you would expect it to be. So your question can be re-interpreted as asking if the following diagram commutes:
$$\require{AMScd}
\begin{CD}
\wedge^2 V @>{a}>> V \otimes V\\
@V{\wedge^2 A}VV @VV{A^{\otimes 2}}V \\
\wedge^2 V @>{a}>> V \otimes V
\end{CD}$$
and this should be clear.
