Let $V$ be a  $2n$-dimensional complex  vector space with base   $\{e_1,\dotsc,e_n,f_1,\dotsc,f_n]\}$ Let $W \subset \wedge^n V$ be the subspace in the exterior product,
with basis vectors
$$
e_{i_1} \wedge \dotsb \wedge e_{i_k} \wedge 
f_{j_1} \wedge \dotsb \wedge f_{j_{n-k}}
$$
where we take all possible indices such that $\{i_1,\dotsc,i_k\} \cup \{j_1,\dotsc,j_{n-k}\}$ is a set partition of $\{1,\dotsc,n\}$. Thus, $W$ is $2^n$-dimensional.

As an example, when $n=2$, we have that $W$ has the following four vectors as basis.
$$
e_1 \wedge e_2, \quad
e_1 \wedge f_2, \quad
e_2 \wedge f_1, \quad
f_1 \wedge f_2
$$

Suppose now that we have a map $T:V \to V$.
It has a natural extension to $\wedge^n V$, (we use $T$ to denote this extension as well) and suppose that $T$ preserves the subspace $W$.
Hence, $T$ is also a linear map from $W$ to $W$.

Suppose furthermore that $T$ is diagonalizable, with eigenvalues $x_1,\dotsc,x_{2n}$.
Then the trace of the map $T:V\to V$ is simply $x_1+\dotsb+x_{2n}$.

It is straightforward to compute the trace of the induced map $T:\wedge^n V \to \wedge^n V$, it is simply $e_n(x_1,\dotsc,x_{2n})$, where $e_n$ denotes the $n$th elementary symmetric function.

**Question I:** How can one express the trace of $T:W \to W$? Is the information given even sufficient?

**Question II:** I am actually only interested in the case 
when $T:V \to V$ is defined as the cyclic shift,
$$
T(e_i) = e_{i+1}, T(e_n)=f_1,
T(f_i) = f_{i+1}, T(f_n)=e_1,
$$
and powers of $T$.
Here, the eigenvalues of $T$  $x_1,\dotsc,x_{2n}$ are simply the roots of $t^{2n}-1=0$.

I think the trace should be $\prod_{j=1}^n (1+\xi^j)$
where $j$ is a primitive $2n$th root of unity, but I cannot really nail down the motivation.