$\newcommand{\IF}{\mathbb{F}}\newcommand{\IC}{\mathbb{C}}$ For simplicity I'm going to assume $q=p$.

Since all elements in $\overline{\IF_p}$ have order $p$, any group homomorphism $\overline{\IF_p}\to\IC^\times$ has either image contained in the group of $p$-th roots of unity $\mu_p$. In other words we can equivalently ask for homomorphisms of additive groups $\overline{\IF_p}\to\mathbb{Z}/p$, i.e. $\IF_p$-vector space homomorphisms $\overline{\IF_p}\to\IF_p$.

Let $F$ be the Frobenius automorphism. Since $F-1: \overline{\IF_p}\to\overline{\IF_p}$ is surjective, we have $\phi\circ F=\phi \iff 0=\phi\circ(F-1) \iff \phi = 0$.

The next best thing is the normalised galois trace $\tau: \bigcup_{(p,m)=1} \IF_{p^m} \to \IF_p$. That is the unique (up to multiplication with a constant) $F$-invariant linear form $\overline{\IF_p}\to\IF_p$.

Every finite field has a one-dimensional space of $F$-invariant linear forms, namely the dual space of $\IF_{p^m} / im(F_{|\IF_{p^m}}-1)$. That is one-dimensional because $\dim im(F-1) = m - \dim ker(F-1) = m-1$. The problem of course is that there is no canonical one and these linear forms do not extend to $F$-invariant linear forms on field extensions of $\IF_{p^m}$ in general (they extend iff the degree of the extension is not divisible by $p$).

Together with $\tau$ you can build an $F$-invariant linear form on $\IF_{p^a} \otimes \bigcup_{(p,m)=1} \IF_{p^m}$ but that is as far as you can go.

$\newcommand{\IF}{\mathbb{F}}\newcommand{\IC}{\mathbb{C}}$ For simplicity I'm going to assume $q=p$.

Since all elements in $\overline{\IF_p}$ have order $p$, any group homomorphism $\overline{\IF_p}\to\IC^\times$ has either image contained in the group of $p$-th roots of unity $\mu_p$. In other words we can equivalently ask for homomorphisms of additive groups $\overline{\IF_p}\to\mathbb{Z}/p$, i.e. $\IF_p$-vector space homomorphisms $\overline{\IF_p}\to\IF_p$.

Let $F$ be the Frobenius automorphism. Since $F-1: \overline{\IF_p}\to\overline{\IF_p}$ is surjective, we have $\phi\circ F=\phi \iff 0=\phi\circ(F-1) \iff \phi = 0$.

The next best thing is the normalised galois trace $\tau: \bigcup_{(p,m)=1} \IF_{p^m} \to \IF_p$. That is the unique (up to multiplication with a constant) $F$-invariant linear form on that field.

Every finite field has a one-dimensional space of $F$-invariant linear forms, namely the dual space of $\IF_{p^m} / im(F_{|\IF_{p^m}}-1)$. That is one-dimensional because $\dim im(F-1) = m - \dim ker(F-1) = m-1$. The problem of course is that there is no canonical one and these linear forms do not extend to $F$-invariant linear forms on field extensions of $\IF_{p^m}$ in general (they extend iff the degree of the extension is not divisible by $p$).

Together with $\tau$ you can build an $F$-invariant linear form on $\IF_{p^a} \otimes \bigcup_{(p,m)=1} \IF_{p^m}$ but that is as far as you can go.