Conjecture: Let $H_n$ be the space of $n\times n$ complex Hermitian matrices and let $\varphi:H_n \to H_n$ be a linear map which preserves determinants: \begin{equation} \det \circ \varphi = \det. \end{equation} Then there exists an $A\in SL(n,\mathbb{C})$ such that $\forall X \in H_n$ we have \begin{equation} \varphi(X)=\pm AXA^{\dagger}. \end{equation} where the minus sign in the RHS can only occur if $n$ is even (and is independent of $X$).
(By the way: this question is inspired by the correspondence between these maps (for $n=2$) with Minkowski space with its invariance under Lorentz transformations, as discussed in https://en.wikipedia.org/wiki/Lorentz_group#Relation_to_the_M.C3.B6bius_group)
Conjecture/"almost theorem": Let $\varphi$ be as in the previous conjecture and suppose moreover that $\phi(I)$ is positive-definite. Then there exists an $A \in SL(n,\mathbb{C})$ such that $\forall X \in H_n$ \begin{equation} \phi(X) = AXA^{\dagger}. \end{equation}
almost complete proof: We can fix an onb $\left\{e_1,...,e_n\right\}$ such that the nondegenerate Hermitian map $\varphi(I)$ becomes a positive, diagonal matrix without any zeroes on the diagonal. By conjugating $\varphi$ with some positive diagonal matrix $C$ in the sense of ($\forall X \in H_n$) \begin{equation} \varphi'(X)=C\varphi(X)C^{\dagger}, \end{equation} (after which $\varphi(:=\varphi')$ still preserves Hermiticity and the determinant) we may assume that $\varphi(I)=I$.
Next we have $\forall \lambda \in \mathbb{R}$ that \begin{equation} \det(\lambda I - X) = \det(\varphi(\lambda I - X)) = \det(\lambda \varphi(I)-\varphi(X)) = \det(\lambda I - \varphi(X)) \end{equation} from which it follows that for any Hermitian $X$ the Hermitian $\varphi(X)$ has the same eigenvalues. Hence there is some unitary matrix $U(X)$ such that \begin{equation} \varphi(X)=U(X)XU^{\dagger}(X)=U^{\dagger}(X)XU(X). \end{equation} To conclude, we must show that $U(X)$ is independent of $X$. In fact, that need not be strictly true: we can take some $U$ such that $\forall X \neq I$: $U(X)=U$ and take any different $V$ such that $U(I)=V$. But I need an argument that the linearity of $\varphi$ allows to choose the function $X \mapsto U(X)$ in a way that it is constant. This seems very plausible to me but I lack a definite verification.
A related question ( https://mathoverflow.net/questions/522/linear-transformation-that-preserves-the-determinant ) on the parallel forum already got a nice answer by Eric Wofsey. Assume that we have proven that the determinant and hermiticity preserving map $\varphi$ takes a form \begin{equation} X \mapsto NXM \end{equation} or \begin{equation} X \mapsto NX^tM \end{equation} (transposition however preserves eigenvalues, multiplicities and ranks (row or column rank, whatever). So I think that $X^t=PXP^{-1}$ in general and $X^{t}=UXU^{\dagger}$ if $X$ is Hermitian. Hence we continue with a map of the form $\varphi:X \mapsto NXM$) where $det(NM)=1$. Let's explore the condition that $\varphi$ preserves Hermitian matrices: Let \left{e_1,...,e_n\right} be an onb and let $P_i$ be the projector matrix which projects on $e_i$. Then we have ($i$ is not being summed over here!) \begin{equation} N_{ai}M_{ib}=\varphi(P_i)_{ab}=\left(\varphi(P_i)_{ba}\right)^*=\left(N_{bi}M_{ia}\right)^*. (A) \end{equation} So, after first fixing some $i$ and subsequently a $b$ such that $N_{bi}\neq 0$ (that is possible because $\det N \neq 0$), we get \begin{equation} M_{ia}^*=\frac{M_{ib}}{N_{bi}^*}N_{ai}=C_i N_{ai} \end{equation} which shows that the i'th column of $N$ is proportional to the complex conjugate of the i'th row of $M$ (going back to (A) we can see that $C_i \in \mathbb{R}$): $N=CM^{\dagger}$ where $C$ is diagonal, real and nonsingular. We can carry on by evaluating $\varphi$ on the following basis of Hermitian matrices: \begin{equation} (P_{ij})_{ab}=\delta_{ia}\delta_{jb}+\delta_{ib}\delta_{ja} \end{equation} \begin{equation} (Q_{ij})_{ab}=i\delta_{ia}\delta_{jb}-i\delta_{ib}\delta_{ja} \end{equation} The requirements that $\varphi(P_{ij})_{ab} = \varphi(P_{ij})_{ba}^*$ and $\varphi(Q_{ij})_{ab} = \varphi(Q_{ij})_{ba}^*$ translate to \begin{equation} (C_jN_{ai}N_{bj}^*-C_j^* N_{bi}^*N_{aj})=(C_i N_{ai}N_{bj}^*-C_i^*N_{bi}^*N_{aj}) \end{equation} and \begin{equation} (C_jN_{ai}N_{bj}^*+C_j^* N_{bi}^*N_{aj})=(C_i N_{ai}N_{bj}^*+C_i^*N_{bi}^*N_{aj}). \end{equation} By subtracting both equations and choosing $a$ and $b$ such that $N_{ia}\neq 0 \neq N_{jb}$ (such indices exist), we can deduce that $C_i=C_j$ such that $C=cI$ is a real multiple of the unit matrix. By defining $A=|c|^{\frac{1}{2}}N$, we see that \begin{equation} \varphi:X\mapsto \pm AXA^{\dagger}. \end{equation}