Equivalent linear representations of finite group over field are equivalent over subfield in characteristic zero

In terms of linear representations
Suppose $$G$$ is a finite group. Suppose $$K$$ and $$L$$ are fields of characteristic zero with $$K$$ a subfield of $$L$$. Suppose $$\varphi_1,\varphi_2:G \to GL(n,K)$$ are linear representations of $$G$$ over $$K$$. We can view $$\varphi_1$$ and $$\varphi_2$$ as linear representations over $$L$$.

The claim is that if $$\varphi_1,\varphi_2$$ are fact about::equivalent linear representations over $$L$$, they are also equivalent linear representations over $$K$$.

In terms of finite subgroups and conjugacy-closed
Suppose $$K$$ and $$L$$ are fields of characteristic zero with $$K$$ a subfield of $$L$$. Then, the general linear group $$GL(n,K)$$ is a finite subgroup-conjugacy closed subgroup inside the general linear group $$GL(n,L)$$. In other words, any two finite subgroups of $$GL(n,K)$$ that are conjugate inside $$GL(n,L)$$ are also conjugate inside $$GL(n,K)$$.

Conjugacy-closed angle

 * General linear group over subfield is conjugacy-closed: This is for elements, possibly of infinite order.

Facts used

 * 1) uses::Character determines representation in characteristic zero, which in turn relies on the uses::orthogonal projection formula which in turn relies on the uses::character orthogonality theorem, which in turn uses uses::Schur's lemma.

Short proof
The proof follows directly from Fact (1). If $$\varphi_1,\varphi_2$$ are equivalent over $$L$$, they in particular have the same character. By Fact (1), they must be equivalent over $$K$$.

What's going on
This superficial-looking proof can obscure what's going on. The real meat of the proof lies in the following aspect of the proof of the character orthogonality theorem: two irreducible representations that are inequivalent over $$K$$ have characters that are orthogonal to each other. However, if they were equivalent over $$L$$, then applying the character orthogonality theorem inside $$L$$, we'd get a nonzero inner product value of the characters. This explains why the statement is true for irreducible characters, and the rest follows by piecing things together.