Determinant of a linear operator restricted to a subspace of $\mathbb{C}^{n\times n}$ Consider $f_B:\mathbb{C}^{n\times n}\to \mathbb{C}^{n\times n}: A \mapsto BAB^+$ where $B$ is some arbitrary matrix in $\mathbb{C}^{n\times n}$ and $B^+$ denotes the conjugate transpose.
By considering $f_B$ as the composition of a left and right multiplication, and taking the standard basis for matrices (e.g. ordered in column major fashion for left multiplication, it is easily shown $\det{f_B}=(\det{B})^n(\det{B^+})^n=|\det{B}|^{2n}$.
Now let $H\subset \mathbb{C}^{n\times n}$ be the subset of hermitian matrices. $H$ is a vector space over the scalar field $\mathbb{R}$, and we can define $g_B:H\to H: A \mapsto BAB^+$ where $B$ is again some arbitrary matrix in $\mathbb{C}^{n\times n}$ (not necessarily hermitian).
I need to show $\det{g_B}=|\det{B}|^{2n}$.
I can see $H$ is a subspace of $\mathbb{C}^{n\times n}$, and $g_B$ is the restriction of $f_B$ to $H$, but of course the determinant of a restriction need not equal that of the full map.
I can also see that $\dim{H}=n^2$ and so $H$ is isomporphic to $\mathbb{R}^{n\times n}$, but I am at a loss as to how I could use this.

In fact, it is not only the case that $g_B$ is the restriction of $f_B$ to $H$, it is also true that $f_B$ is the unique extension of $g_B$ to a complex-linear map. From this, we can deduce that they must have the same determinant.
To put this another way: suppose that we have a basis $\{H_1,H_2,\dots,H_{n^2}\}$ for the real-linear space of Hermitian matrices. Note that this set also forms a basis for $\Bbb C^{n \times n}$ under complex linear combinations. Thus, the matrix of $A \mapsto BAB^+$ relative to this basis is the matrix of both $f_B$ and $g_B$. So, the two maps must have determinant equal to the determinant of this matrix.
