Let $A\colon=(a_{jk})$ be the matrix of $v\in\Hom(E)$ with respect to the basis $e_1,\ldots,e_n$. Then $B=(b_{jk})\colon=U^{-1}AU$ is the matrix of $v\in\Hom(E)$ with respect to the basis $b_1,\ldots,b_n$.
Since the inverse $U^{-1}=(u^{jk})$ of $U=(u_{jk})=(e_j^*(b_k))$ is given by $u^{jk}=b_j^*(e_k)$, we get
$$
b_{jk}
=b_j^*(v(b_k))
=b_j^*\Big(\sum_le_l^*(b_k)v(e_l)\Big)
=\sum_{l,m}e_l^*(b_k)a_{ml}b_j^*(e_m)
=\sum_{l,m}u_{lk}a_{ml}u^{jm}
=\sum_{l,m}u^{jm}a_{ml}u_{lk}~.
$$