Let $q = 2^n$ with $n$ even. The finite field $\mathbb F_q$ then has a (unique) automorphism $\theta$ of order two, since the Galois group of $\mathbb F_q/\mathbb F_2$ is cyclic of order $n$. The corresponding fixed field $\operatorname{Fix}(\theta) = \{ x \in \mathbb F_q \mid \theta(x) = x \}$ is then isomorphic to the finite field $\mathbb F_{2^m}$ where $m = n/2$. Of course, the degree of $\mathbb F_q/\operatorname{Fix}(\theta)$ equals 2 and so there is a decomposition of vector spaces of the form $\mathbb F_q \cong \mathbb F_{2^m} \oplus \mathbb F_{2^m}$ that depends on a choice of basis.
My question is now as follows: Is there a natural way to choose a complement of the $\mathbb F_{2^m}$-subvector space $\operatorname{Fix}(\theta)$ in $\mathbb F_q$? For example, one can show that $$ \operatorname{Fix}(\theta) = \{ \theta(x) + x \mid x \in \mathbb F_q \} = \{ x\cdot\theta(x) \mid x \in \mathbb F_q \} $$ So given an element $x \in \mathbb F_q$ one might first write $$ x = (\theta(x) + x) + \theta(x) = x\theta(x) + (x + x\theta(x)) $$ However, whereas the first summand lies in $\operatorname{Fix}(\theta)$, the second components will not span a complement to $\operatorname{Fix}(\theta)$.