Determinant and trace as conjugations? For real matrices $A$ it holds that $$\det\,\big(e^A\big)=e^{\mathrm{tr}\,A}$$ so we can write $$\mathrm{tr}=(\exp)^{-1}\circ \;\det\;\circ\;(\exp).$$ Is this interpretation of trace as the "conjugate" of the determinant under the exponential map used anywhere, or useful for anything? It seems neat but I have not come across it before.
Edit: note that the $(\exp)$ on the left (of which an inverse is taken) is the natural logarithm $\ln$ for real numbers, not for matrices (I think!) because it is applied after finding the determinant.
 A: In the theory of matrix Lie groups the exponential map represents the passage from Lie algebra to its Lie group. Lie algebra 'linearizes' the group in the neighborhood of the identity matrix $I$ (in technical terms it is the tangent space at $I$). The special linear group $SL(V)$ on a vector space $V$ consists of matrices with determinant $1$, that determinant is conjugate to the trace implies immediately that the corresponding Lie algebra $\mathfrak{sl}(V)$ consists of matrices of trace $0$. Note also that for non-commuting matrices $e^{A+B}\neq e^Ae^B$, however we still have $\det e^{A+B}=\det e^A\,\det e^B$ due to conjugacy with the trace.
There is a different kind of exponential map in Riemannian geometry related to geodesics. Although things are much more complicated than with matrices conceptually determinant corresponds to volume, and there are limit formulas that express local deviation from Euclidean volume in terms of Ricci and scalar curvatures, which are 'traces' of the Riemannian curvature.
A: It is absolutely useful in the study of Lie groups, where the exponential map takes us from a Lie algebra to the corresponding Lie group, which allows for the study of these groups "at the algebra level".
I'm not sure if your particular property ends up being useful, but one certainly uses the fact that eigenvalues are mapped as $\lambda \mapsto e^{\lambda}$.
