In his The Symmetric Group, Bruce Sagan proves the so-called character relations of the first kind:
Let $\chi$ and $\psi$ be irreducible characters of a group $G$. Then $\langle \chi,\psi\rangle = \delta_{\psi,\chi}$.
The inner product of characters is defined as $\langle \chi,\psi\rangle= \frac{1}{|G|} \sum_{g\in G} \chi(g)\psi(g^{-1})$.
The prove starts by considering the matrix representations $A$ of degree $d$ and $B$ of degree $f$ corresponding to $\chi$ and $\psi$ respectively. Let $X=(x_{i,j})$ be a $d\times f$-matrix of indeterminates $x_{i,j}$ and define the matrix $$ Y = \frac{1}{|G|}\sum_{g\in G} A(g)X B(g^{-1}).$$ Then we prove that $A(h)Y=YB(h)$ for all $h\in G$. By Schur's lemma, this implies that $Y=0$ if $A\not\cong B$ and $Y=cI_d$ if $A\cong B$.
We first suppose that $\chi\ne\psi$, which implies $A\not\cong B$, and show that $\langle \chi,\psi\rangle=0$. Next up comes the part where I'm stuck: assume that $\chi=\psi$. The proof then states that since we're only interested in the character values, we might as well take $A=B$. I don't understand why one can make this assumption. It seems too restrictive.
Per assumption we have that $\chi(g)=\psi(g)$ for all $g\in G$, hence $\operatorname{tr}A(g)=\operatorname{tr}B(g)$ for all $g\in G$. It seems like it's even too restrictive to ask that $A$ and $B$ have the same diagonal entries, let alone all elements being the same...
Note: at this point in the notes, we have only shown that if $X$ is a matrix representation of $G$ with character $\chi$ and $Y$ with character $\psi$ then $X\cong Y\Rightarrow \chi =\psi$. The other implication is also true, but cannot be used to argue why $A=B$ is a valid assumption.
It's true that if $\chi=\psi$, then $A\cong B$ are equivalent reps, which means $B(g)=CA(g)C^{-1}$ for some intertwiner $C$, i.e. the $A(g)$s and $B(g)$s are simultaneously conjugate. This does not imply $A=B$, or that they have the same diagonal entries, but it does imply $\mathrm{tr}A(g)=\mathrm{tr}B(g)$.
You don't actually need to know any of this, though. All you need to know is that $A=B$ implies $\mathrm{tr}A=\mathrm{tr}B$, because that's all the proof is using. Keep in mind we are not starting with $A,B$ and then defining $\chi,\psi$ from them, but the other way around: we start with $\chi,\psi$ and then pick $A$ and $B$ to have those characters. If the chars are the same, we might as well pick $B$ to be $A$ as well.