Given a finite dimensional (real) vector space $V$, and two non-degenerate bilinear forms $(\cdot,\cdot)\_1$ and $(\cdot,\cdot)\_2$, one can use a basic linear algebra argument to show that there exists a unique invertible linear map $F:V \to V$ such that 
$$
(F(\cdot),\cdot)\_1 = (\cdot,\cdot)\_2.
$$
Now if we also assume that $V$ carries a representation of a group $G$, with respect to which both $(\cdot,\cdot)\_1$ and $(\cdot,\cdot)\_2$ are $G$-equivariant, which is to say 
$$
(v.g,w.g)\_1 = (v,w)\_1, ~~~~~~ (v.g,w.g)\_2 = (v,w)\_2, ~~~~~~~ \text{ for all } v,w \in V, g \in G.
$$
In this case will the map $F$ always be $G$-equivariant, i.e. will we always have
$$
F(v.g) = F(v).g,
$$
or do there exist situations where this equality will not hold? Another way to express the question is: Given an equivariant non-degenerate bilinear form $(\cdot,\cdot)$, and an invertible equivariant map $F$, one can produce the new equivariant non-degenerate bilinear form $(F(\cdot),\cdot)$, but can all equivariant non-degenerate bilinear forms be produced in this way from $(\cdot,\cdot)$? In other words can we classify all equivariant non-degenerate bilinear forms in terms of a certain bilinear form $(\cdot,\cdot)$ and all equivariant invertible maps $L$.



Classifying all Equivariant Bilinear Forms on a Finite-Dimensional Module
This is true: Write $(v,w)\_i = \langle A\_i(v),\rangle$ where $A\_i:V\to V^\ast$ is invertible and $\langle\quad,\quad\rangle:V^\ast\times V \to K$ is the duality pairing. $(\quad,\quad)\_i$ is $G$-invariant iff $A\_i$ is $G$-equivariant, and $A\_1\circ F=A\_2$, or $F=A\_1^{-1}\circ A\_2$ is then $G$-equivariant too. 


