Let $\eta \in \operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $\eta$ with respect to the basis $(e_1, \dots e_n)$. Let $f_i = \sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) \in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 \dots f_n)$.
$\underline{My \space attempt:}$
$$\eta(f_i) = \eta\left(\sum_{j=1}^n p_{i,j}e_j \right) = \sum_{j=1}^n \eta(p_{i,j} e_j) = \sum_{j=1}^n p_{i,j} \eta(e_j)$$
$$=\sum_{j=1}^n p_{i,j} \left( \sum_{k=1}^n a_{i,k} e_k \right) = \sum_{j=1}^n \sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$
$$=\sum_{j=1}^n\sum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$
From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!
I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.
We could finish your proof as follows: $$ \eta(f_i) = \eta \left( \sum_{j=1}^n p_{ij} e_{j}\right) = \sum_{j=1}^n p_{ij} \eta(e_{j})\\ = \sum_{j=1}^n p_{ij} \left(\sum_{k=1}^n a_{jk}e_{k}\right) = \sum_{k=1}^n \sum_{k=1}^n p_{ij}a_{jk} e_k \\= \sum_{j=1}^n \sum_{k=1}^n p_{ij}a_{jk} \left(\sum_{\ell=1}^n p^{-1}_{k\ell}f_{\ell}\right) = \sum_{j=1}^n \sum_{k=1}^n \sum_{\ell=1}^n p_{ij}a_{jk}p^{-1}_{k\ell} f_\ell $$ Now, it suffices to note that $\sum_{j=1}^n \sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{k\ell}$ is indeed the $(\ell,i)$ of the matrix $PAP^{-1}$.