In the Artin's book on Algebra, the author stated a theorem (Ch.9, Thm. 2.2):
"A finite subgroup $G$ of $GL(n,\mathbb{C})$ is conjugate to a subgroup of $U(n)$.
Here, $U(n)$ is the unitary group, i.e. if $\langle \,\,, \rangle$ is the standard Hermitian inner product on $\mathbb{C}^n$ given by $\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n)\rangle=\sum_{}{x_i\bar{y_i}}$ then $U(n)=\{A\in GL(n,\mathbb{C}) \colon \langle Av,Aw\rangle = \langle v,w\rangle,\forall v,w\in \mathbb{C}^n \}$.
The proof is: there is a $G$-invariant Hermitian inner product $\langle\,\, , \rangle_1$ on $\mathbb{C}^n$, and consider an orthonormal basis $B_1$ w.r.t this form. If $P$ is the matrix of transformation which changes standard basis to $B_1$, then $PGP^{-1}\leq U(n)$
Question: In the last statement in the proof, the author says that $PGP^{-1}$ is subgroup of unitary group; but, this unitary group is corresponding to the Hermitian inner product $\langle \,\,, \rangle_1$, i.e. it is the group of linear transformations which preserves the inner product $\langle \,\,, \rangle_1$. Why $PGP^{-1}$ should be subgroup of $U(n)$ which is the group of linear transformations which "preserves" the standard Hermitian inner product $\langle \,\,,\rangle$ on $\mathbb{C}^n$?