I have a question about the structure of $G$-invariant bilinear forms. Let $G$ be an arbitrary finite group and $\mathbb{F}_q$ a finite field such that $2|G|$ is not divisible by the characteristic of the field. Let $V$ be an irreducible $\mathbb{F}_qG$-module for which the dimensions of the spaces of $G$-invariant symmetric bilinear forms and skew-symmetric bilinear forms are both of dimension $>0$ (in particular the module is self-dual).
Now we can consider the decomposition of $V$ into absolutely irreducible constituents, which I group as follows: $$\overline{\mathbb{F}_q}\otimes_{\mathbb{F}_q} V \cong \bigoplus_{i=1}^s V_i^{\alpha_i} \oplus \bigoplus_{j=1}^t W_j^{\beta_j} \oplus \bigoplus_{\ell=1}^r (X_\ell \oplus X^* _\ell)^{\gamma_j}$$ Here the $V_i$ have a one-dimensional space of $G$-invariant symmetric forms each, the $W_j$ have a one-dimensional space of skew-symmetric $G$-invariant forms each and the $X_\ell$ only have $0$ as a $G$-fixed form, which means that they are not self-dual and the dual $X_\ell^*$ occurs in the decomposition with the same multiplicity as $X_\ell$. I think it is not unusual to call the $V_i$ orthogonal components and the $W_j$ symplectic components.
Now, based on some examples I have looked at and some "evidence" I have gathered from some computer code, I think that in the presence of both symmetric and skew-symmetric forms on the $\mathbb{F}_q G$-module $V$ there will be neither any orthogonal nor any symplectic constituents, or in other words $s=t=0$, but I have absolutely no idea how to prove this. Another way of looking at this might be that the existence of both orthogonal and symplectic constituents in the decomposition over the algebraic closure (or a splitting field) implies that $V$ is not irreducible, but this approach didn't get me very far either... One idea in that case might be to show that $\bigoplus_{i=1}^s V_i^{\alpha_i}$, i.e. the submodule of all orthogonal constituents with multiplicities, is already realizable over $\mathbb{F}_q$, but I don't really have an idea how to prove this (if it is even true...).
So, I am looking for help with regard to this question. Also, and this occured to me just a few minutes ago, might this statement still be true without the assumption that the field be finite (of course still maintaining the condition that the group algebra be semi-simple and that there exist a distinction between symmetric and skew-symmetric, i.e. $\mathrm{char}(\mathbb{F})\neq 2$)?
I am looking forward to any kind of hint, because this problem really doesn't seem like it should be so difficult...
The statement above is correct, as can be seen by applying the functor $$F ~:~\overline{\mathbb{F}_q}G-\mathrm{mod} \to \overline{\mathbb{F}_q}G-\mathrm{mod}$$ which sends a module $V$ to ${^\sigma V}$, which is the same module, but with scalar multiplication twisted by $\sigma ~:~ x\mapsto x^q$, the Frobenius automorphism of $\overline{\mathbb{F}_q}$ with fixed field $\mathbb{F}_q$.
The following facts then imply the statement.