Given a representation of a finite group: $\rho : G \to \text{GL}(V)$, there exists a unitary representation $\tau$ which is isomorphic to $\rho$. I came across Weyl's trick, where you redefine the inner product as follows $$\langle v,w \rangle = \frac1{|G|}\sum_{g \in G} \langle \rho(g)v, \rho(g)w \rangle_0,$$
where $\langle\ ,\rangle_0$ is the usual inner product. I see the idea of this trick since this new inner product is preserved under $\rho$ i.e. $\langle v,w \rangle = \langle \rho(g')v,\rho(g')w \rangle$ for all $g'\in G$.
I am not familiar with the idea of changing the definition of the inner product to achieve unitarity. As I understand Weyl's trick, the matrices in the set $\{\rho(g):g\in G\}$ and $\{\tau(g):g\in G\}$ are actually identical but the inner product definition is what makes the $\tau(g)$ matrices unitary? Is there a way to translate this back to the usual picture where I use the standard inner product (and therefore a fixed definition of unitarity) and the matrices of the different representations are not identical?
Sorry if there is some vagueness/incorrectness in the question - I am new to group theory.
If you want to be technical, the way you've defined it, $\rho(g)$s are not matrices, they are functions. Of course, given a choice of ordered basis for $V$, we can write vectors $v\in V$ as coordinate vectors and linear transformations $A\in\mathrm{GL}(V)$ as matrices. What matrices these functions get represented as though depends on which choice of ordered basis (i.e. coordinates) one picks.
I wouldn't say there is a "usual inner product" on an arbitrary vector space $V$. There is just an arbitrary inner product that one selects or assumes is given. Note that even having an inner product does not provide enough information to represent the functions $\rho(g)$ as matrices; an inner product does not provide for an ordered basis, which is what is needed for that.
If you assume $\rho$ is a matrix representation from the get-go, that is if you define $\rho:G\to GL_n (\mathbb{C})$ as a representation of $G$ on the vector space $\mathbb{C}^n$, there is no guarantee the matrices $\rho(g)$ are unitary, i.e. there is no guarantee they preserve the standard inner product on $\mathbb{C}^n$,
$$ \langle x,y\rangle=\overline{x_1}y_1+\cdots+\overline{x_n}y_n. $$
We can define the $G$-averaged inner product $\langle x,y\rangle_G$ by the unitary trick:
$$ \langle x,y\rangle_G = \frac{1}{|G|} \sum_{g\in G} \langle \rho(g)x,\rho(g)y\rangle. $$
Then $\rho(g)$s my not preserve the standard inner product $\langle x,y\rangle$, but they do preserve $\langle x,y\rangle_G$, which when written out will have a different formula than $\langle x,y\rangle$ does.
As I said, the new inner product $\langle x,y\rangle_G$ does not automatically provide a way to rewrite linear operators as matrices. However, if one arbitrarily chooses an ordered unitary basis with respect to $\langle x,y\rangle_G$, we can rewrite the matrices $\rho(g)$ as the matrices $\tau(g)$ using a change-of-basis matrix $M$. Explicitly, these matrices are given by $\tau(g)=M^{-1}\rho(g)M$, where the columns of $M$ are the vectors of the ubasis we chose.
These $\tau(g)$s will be unitary matrices and will preserve the original, standard inner product $\langle x,y\rangle$, because $\tau(g)$'s action on the standard basis corresponds to $\rho(g)$s action on the chosen basis, and $\rho(g)$s action on the chosen basis (unitary with respect to $\langle-,-\rangle_G$) is unitary. Explicitly, because the columns of $M$ are unitary with respect to $\langle-,-\rangle_G$ we can write $Mu$ and $Mv$ as linear combinations of $M$'s columns in order to verify $\langle Mu,Mv\rangle_G=\langle u,v\rangle$. Therefore,
$$ \begin{array}{ll} \langle \tau(g)u,\tau(g)v\rangle & = \langle M^{-1}\rho(g)Mu,M^{-1}\rho(g)Mv\rangle \\ & = \langle \rho(g)Mu,\rho(g)Mv\rangle_G \\ & = \langle Mu,Mv\rangle_G \\ & = \langle u,v\rangle \end{array} $$
so $\tau(g)$ is unitary with respect to $\langle-,-\rangle$.