I want to show that for each unitary representation $\pi: G \to GL(\mathbb{C}^n)$, the matrix $A \in GL_n(\mathbb{C})$ "corresponding" to $\pi$ is also unitary. So I suppose $\pi$ is unitary with respect to the usual inner product on $\mathbb{C}^n$. This means
$$\forall \, g \in G, \forall \, v,w \in \mathbb{C}^n, \langle \pi_g(v), \pi_g(w) \rangle = \langle v , w \rangle = v \cdot \overline{w}$$
Fix $g \in G$ and let $A \in GL_n(\mathbb{C})$ be the matrix corresponding to $\pi_g$. Then this means
$$\langle Av , Aw \rangle = \langle v , w \rangle = v \cdot \overline{w}$$
But then
$$\langle Av, Aw \rangle = Av \cdot \overline{Aw} = ???$$
In essence, I seek to arrive at the conclusion that $A^{-1} = A^* = \overline{A^\top}$, which is exactly what it means for $A$ to be unitary.
Am I off to the right start? How can I manipulate the ending line of my proof to conclude that $A$ is unitary?
The (hermitian) dot product is given by the matrix multiplication: $$ \langle v,w\rangle = v^t \overline{w} $$ where $v,w$ are column vectors ($n\times 1$ matrices). Now you can combine with the fact that $(Av)^t=v^tA^t$.
Edit (response to comment):
You're almost there! You have $A^t\overline{Aw} = \overline w$. Since this is true for all vectors $w$, you have that $A^t\overline A = \text{id}$. (In terms of matrices, apply the above to each basis vector $\overline{w}=e_i$ and you'll get that each column of the two matrices $A^t\overline A$ and $\text{id}$ are equal.)
Finally take the conjugate of the equation to get $A^*A = \text{id}$.