There is a detail that I had the indicies wrong when I wrote it on the board. This sort of linear algebra is not hard, but it is all about getting the accounting right and being organized.

I said let ${\mathcal B} = \{ v_1, v_2, \cdots, v_n \}$ and ${\mathcal C} = \{ w_1, w_2, \cdots, w_n \}$ be bases for a $G$ module $V$. Then there are coefficients, $$g(v_i) = \sum_{j=1}^n a_{ji} v_j$$ $$g(w_i) = \sum_{j=1}^n b_{ji} w_j$$ and $$v_i = \sum_{j=1}^n t_{ji} w_j$$

If we first compute the action of $g$ on $v_i$ and expand in the basis ${\mathcal C}$, then we have $g(v_i) = \sum_{j=1}^n a_{ji} v_j = \sum_{j=1}^n \sum_{k=1}^n a_{ji} t_{kj} w_k$ If instead we compute the action of $g$ on $v_i$ by expanding in the basis $\mathcal C$ followed by the action of $g$ on the $\mathcal C$ basis we have $g(v_i) = \sum_{j=1}^n t_{ji} g(w_j) = \sum_{j=1}^n \sum_{k=1}^n t_{ji} b_{kj} w_k$

Now let $T = [t_{ji}]_{1 \leq i,j \leq n}$ be the matrix of coefficients for the change of basis matrix between the $\mathcal B$ basis and $\mathcal C$ basis

The coefficient of $w_k$ in $g(v_i)$ using the first of these two equations is equal to the $(k,i)$ entry in the matrix $T \cdot X_{\mathcal B}(g)$ where $X_{\mathcal B}(g) = [ a_{ji} ]_{1 \leq i,j \leq n}$.

The coefficient of $w_k$ in $g(v_i)$ using the second of these two equations is equal to the $(k,i)$ entry in the matrix $X_{\mathcal C}(g) \cdot T$ where $X_{\mathcal C}(g) = [ b_{ji} ]_{1 \leq i,j \leq n}$.

Since these two quantities must be equal for the action to be consistent on the the bases, we must have $T \cdot X_{\mathcal B}(g) = X_{\mathcal C}(g) \cdot T$