Basis for complex representation with symmetric bilinear form For a finite group $G$, if a complex irreducible representation $\rho:G\to\mathrm{GL}(V)$ has a $G$-invariant nondegenerate symmetric bilinear form $B:V\times V\to\mathbb{C}$, then there is a basis of $V$ with respect to which each matrix of each $\rho_g$ has only real entries.  My question is whether there is some direct way of seeing this or producing such a basis.
First, the reason I know this is true is through the Schur indicator.  Since there is a $G$-invariant nondegenerate symmetric bilinear form, the representation $S^2V^*$ is nontrivial, and since $\hom_G(V,V^*)=V^*\otimes V^*=S^2V^*\oplus \Lambda^2V^*$, this implies $V$  and $V^*$ are isomorphic representations.  In particular, taking $V^{\mathbb{R}}$ to be real representation obtained from $\mathbb{C}$ as an algebra over $\mathbb{R}$, $(\chi_{\rho^{\mathbb{R}}},\chi_{\rho^{\mathbb{R}}})=4$ since  $\chi_{\rho^{\mathbb{R}}}=\chi_{\rho}+\chi_{\rho^*}$.  The possibilities for $\operatorname{End}_G(V^{\mathbb{R}})$ with this dimension are either $\mathbb{H}$ or $M_2(\mathbb{R})$, the algebra of $2\times 2$ matrices over $\mathbb{R}$, and since $\rho$ does not admit a skew-symmetric $G$-invariant bilinear form, it cannot be $\mathbb{H}$.  Then, $M_2(\mathbb{R})$ has projection matrices, so we can project onto a $V$ in $V^{\mathbb{R}}$.  Since the matrices of $\rho^{\mathbb{R}}$ with respect to any basis have real entries, we can obtain a basis of the original $V$ where the matrices have real entries.
Second, it is not enough (as far as I can tell) to simply take some orthonormal basis of $V$ with respect to $B$.  If $e_1,\dots,e_n$ are such an orthonormal basis, and if $A_g$ is the matrix of $\rho_g$ with respect to this basis, then $G$-invariance gives $A_{g^{-1}}=A_g^T$, so $A_g^TA_g=I_n$ for all $g$, which means the matrices are complex orthogonal matrices.  If it were true that there is a $G$-invariant Hermitian form for which $e_1,\dots,e_n$ are orthogonal, then $A_g^*=A_g^T$, implying $A_g$ has real entries.
 A: Let $\langle-,-\rangle:V\times V\to\mathbb{C}$ denote a $G$-invariant Hermitian form on $V$, which is $\mathbb{C}$-linear in the second variable and $\mathbb{C}$-antilinear in the first variable.  Then the map $V\to V^*$ given by $v\mapsto\langle v,-\rangle$ is a $\mathbb{C}$-antilinear $G$-invariant isomorphism (of vector spaces).  We can then compose this with the $\mathbb{C}$-linear $G$-invariant isomorphism $V^*\to V$ coming from the $G$-invariant nondegenerate symmetric form $(-,-)$.  The composition of these two maps, which I will denote $\tilde{\sigma}:V\to V$, is then a $G$-invariant $\mathbb{C}$-antilinear automorphism of $V$.
Now $\tilde{\sigma}^2:V\to V$ will be a $G$-invariant $\mathbb{C}$-linear automorphism of $V$, and so by Schur's lemma is multiplication by some scalar $\lambda\in\mathbb{C}$.  Define $\sigma:=\frac{1}{\sqrt{\lambda}}\tilde{\sigma}$, and then $\sigma$ is a $G$-invariant $\mathbb{C}$-linear involution.  It follows that $V^\sigma$, the fixed points of $\sigma$, is a real vector space with a $\mathbb{R}$-linear $G$-action such that $V^\sigma\otimes_\mathbb{R}\mathbb{C}\cong V$.  Choosing a basis of $V^\sigma$ as a real vector space gives the desired basis of $V$ as a complex vector space (i.e. the matrices $\rho_g$ will be real with respect to this basis).
