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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 $L$.

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up vote 2 down vote accepted

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.

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