Let $X$ be a scheme over a finite field $\mathbb{F}_q$ and let $F : X \to X$ be the absolute Frobenius morphism. If $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module, then there is a natural isomorphism $F^*(\mathcal{L}) \cong \mathcal{L}^{\otimes q}$.

>Is there also a general formula for $F^*(\mathcal{M})$ if $\mathcal{M}$ is a locally free $\mathcal{O}_X$-module of given rank $d$?

Of course, we have $F^*(\mathcal{M}) = \mathcal{M} \otimes_{\mathcal{O}_X} F_*(\mathcal{O}_X)$ (since $F$ is affine), but I would like to have a formula which is independent of the quasi-coherent algebra $F_*(\mathcal{O}_X)$ (unless you can describe this algebra via generators and relations), but only depends on $\mathcal{M}$ and uses the usual operations on quasi-coherent modules ($\otimes$, $\ker$, $\mathrm{coker}$, $\oplus$, $\mathrm{Sym}^n$, $\Lambda^n$, $\dotsc$).

For $q=2$ I have found that $F^*(\mathcal{M})$ is the kernel of the homomorphism $\mathrm{Sym}^2(\mathcal{M}) \twoheadrightarrow \Lambda^2(\mathcal{M})$, $m \cdot n \mapsto m \wedge n$. In general, there is an embedding$$F^*(\mathcal{M}) \to \mathrm{Sym}^q(\mathcal{M})$$ which maps $m \otimes 1 \mapsto m^q$.

My question may be also phrased as follows: On the $K$-theory of $X$, $F^*$ induces the Adams operation $\psi^q$. Hence, I ask for a bundle representative  of the Adams operation $\psi^q$. (Not just a formal difference of bundles.)