Let $K$ be a field. Let $V$ be a finite-dimensional $K$-vector space; and let $f$ be an automorphism of $V$. Assume that $f$ is diagonalisable over some extension of $K$. Form the exterior algebra $\bigwedge^{*} V$. The automorphism $f$ induces an automorphism $\bigwedge^{*} f$ of $\bigwedge^{*} V$.

Now comes one technical assumption: assume that $f$ generates a (connected) torus in $\mathrm{GL}(V)$.

Can one reconstruct $f$ (up to conjugation) from $\bigwedge^{*} f$, viewing $\bigwedge^{*} V$ as $K$-vector space (i.e., forgetting the graded structure and the algebra structure)?

Observation: it suffices to prove this for algebraically closed fields.

In my specific case, I know slightly more, because I still know the $\mathbb{Z}/2\mathbb{Z}$-grading on $\bigwedge^{*} V$. In other words, I know $\bigwedge^{\text{even}} f$ and $\bigwedge^{\text{odd}} f$.

**Motivation.** I have a system of Galois representations $H_{\ell}$ , and I know that $\bigwedge^{\text{even}} H_{\ell}$ and $\bigwedge^{\text{odd}} H_{\ell}$ form compatible systems of weight $0$. Here compatibility means: the characteristic polynomials of Frobenii have coefficients in $\mathbb{Q}$, are independent of $\ell$, and their roots are $q$-Weil numbers. I would like to deduce that the $H_{\ell}$ form a compatible system themselves.
*Notably, my representations are motivic: they are subrepresentations of $l$-adic cohomology groups. So I know that eigenvalues of Frobenius are Weil numbers; but I do not know a priori that the characteristic polynomials have coefficients in $\mathbb{Q}$.*

**Rephrasing in primary school terms.** Let $S$ be a finite multiset of complex numbers. (I assume your primary school treats basic set theory and complex numbers.) Define the $i$-th exterior multiset
$$\bigwedge^{i} S = \big\{ \prod_{x \in X} x\, \big|\, X \subset S, \#X = i \big\}.$$
Put $\bigwedge^{\text{even}} S = \bigcup_{k \ge 0} \bigwedge^{2k} S$,
and $\bigwedge^{\text{odd}} S = \bigcup_{k \ge 0} \bigwedge^{2k + 1} S$ (as multisets).

The technical condition stated above translates to: assume that all roots of unity in $S$ are equal to $1$.

Do the multisets $\bigwedge^{\text{even}} S$ and $\bigwedge^{\text{odd}} S$ determine the multiset $S$?