Linear maps inducing identity on exterior square Let $V$ be $n$-dimensional space over a field $F$. Let $V\times V\rightarrow \wedge^2 V$ be the natural map. Thus, if $\{v_1,v_2,\ldots, v_n\}$ is a basis of $V$ then $\{v_i\wedge v_j \,\,|\,\, 1\le i<j\le n\}$ is a basis of $\wedge^2V$.

Q. What are the linear maps $\varphi:V\rightarrow V$ which induce identity map on $\wedge^2(V)$? Are such maps well-studied?

Here induced maps on $\wedge^2(V)$ are naturally defined by $T(v\wedge w)=T(v)\wedge T(w)$. 

If necessary, assume that $F$ is of characteristic different from $2$.
 A: I think, only identity linear map of $V$ induces identity on $\wedge^2 V$ if $\dim V\ge 3$. 
If $\dim V=2$, then the group ${\rm SL}(V)$ is the desired set of linear maps inducing identity on $\wedge^2 V$. 
Proof for $n\ge 3$:
Let $T:V\rightarrow V$ be such that induced map on $\wedge^2(V)$ is identity. 
(1) Consider $v,w$ independent in $V$. Then $T(v\wedge w)=v\wedge w$. It is easy to show that $T$ takes span$\{v,w\}$ to itself. 
(2) Let $u$ be arbitrary non-zero vector. Then choose $v,w$, independent vectors such that $\dim$ span$\{u,v\}=2=\dim$ span$\{u,w\}$ and $u,v,w$ are independent.
By (1), $T$ takes span$\{u,v\}$ to itself, and span$\{u,w\}$ to itself; and hence their intersection to itself. This means, $T(u)=\lambda u$ for some $\lambda\in F$. 
From this, it is easy to show that $T$ 
(i) $T$ is scalar matrix 
(ii) The scalar matrix must be identity (due to $T(v\wedge w)=v\wedge w$ for all $v,w\in V$).

I hope this is correct.
Edit (After comment by darij grinberg:) (ii) is incorrect; the scalar matrix is $\pm $ identity. So for $\dim V\ge 3$, $\pm I$ are the only maps inducing identity on $\wedge^2 V$.
