$FG$-modules $V$ and $W$ isomorphic and its properties

Question

I am reading the textbook: "Representations and Characters of Group (G. James & M. Liebeck) p.64"

There is a fact:

The $FG$-modules $V$ and $W$ are isomorphic if and only if there are a basis $\mathcal{B}_1$ of $V$ and a basis $\mathcal{B}_2$ of $W$ such that $$[g]_{\mathcal{B}_1}=[g]_{\mathcal{B}_2} \ \ \ \ \forall g\in G$$

Note: $F$ is a field, $G$ is a finite group.

The proof goes as the following:

($\Rightarrow$)

Suppose $\vartheta$ is an $FG$-isomorphism from $V$ to $W$, and let $v_1.\ldots,v_n$ be a basis $\mathcal{B}_1$ of $V$; then $v_1\vartheta,\ldots,v_n\vartheta$ is a basis $\mathcal{B}_2$ of $W$. Let $g\in G$. Since $(v_ig)\vartheta=(v_i\vartheta)g$ for each $i$, it follows that $[g]_{\mathcal{B}_1}=[g]_{\mathcal{B}_2}$.

My question is:

I know $(v_ig)\vartheta=(v_i\vartheta)g$ is from homomorphism of $V$ and $W$. However, how to go from $(v_ig)\vartheta=(v_i\vartheta)g$ to $[g]_{\mathcal{B}_1}=[g]_{\mathcal{B}_2}$?

Note:
1. A function $f:V\rightarrow W$ is $FG$-homomorphism if $f$ is a linear function and $$(vg)f = (vf)g \ \ \ \ \ \ \forall v\in V, g\in G$$
2. $vg$ is a group action of $g$ on $v\in V$.

Answer 1

If for all $i$, $$v_ig=\sum_ja_{ij}v_j,$$ then $v_ig\theta=v_i\theta g$, so $$(\sum_ja_{ij}v_j)\theta=v_i\theta g,$$ or $$ \sum_ja_{ij}(v_j\theta)=(v_i\theta) g,$$ and so you can see that $\{a_{ij}\}$ is the matrix of $g$ with respect to $\{v_i\}$ and with respect to $\{v_i\theta\}$.