How can one define a kind of "determinant" on a reduced group $C^*$ algebra? Let $A$ be a unital $C^*$-algebra which is equipped with a faithful trace $T$. In particular we may consider $A=C^*_{\text{red}} (G)$ for some discrete group $G$. We consider the following differential equation on $A$:
$$Z'=Z^2-Z.\tag{*}\label{star1}$$
(For $A=M_n(\mathbb{C})$ one can easily check that $$D'=D(T-n)\tag{**}\label{star2}$$ where $D$, $T$ are the standard determinant and trace respectively and $D'$ is the derivative of $D$ along solutions of \eqref{star1}. Note that $n$ in \eqref{star2} can be regarded as $\operatorname{trace}(I_n)$. We will modify this $n$ to $1$ in the case of normalized trace.
In fact "determinant" is the unique analytic function on $M_n(\mathbb{C})$ satisfying the  equation \eqref{star2} with initial condition $D(I_n)=1$.)
We try to generalize this situation of matrix algebra to a $C^*$-algebra $A$ with a faithful normalized trace $T$. So we consider the following modified differential equation:$$D'=D(T-1)\tag{***}\label{star3}$$
where the unknown $D$ is a function on $A$ and $T$ is a normalized trace. Moreover $D'$ is the derivative of $D$ along solution of \eqref{star1}.

What can be said about existence of a global solution $D$ for \eqref{star3} with initial condition $D(1)=1$? Does such a solution $D$ satisfy the multiplicativity condition $D(ab)=D(a)D(b)$? Is $D^{-1}(0)$ equal to the set of all non invertible elements?
As a motivation for the later question, we note that the group of invertible elements is flow-invariant under system \eqref{star1} (see On differential equation $Z'=Z^2-Z$ on a $C^*$ algebra).

If there are no some complete answers to the above questions for an arbitrary algebra with a faithful normalized trace,  what would be the answer of those questions in the particular case $A=C^*_{\text{red}} (G)$? For which kind of groups the answer to the above questions are known?
Remark: We conclude that "determinant" as a function on matrix algebra can be dynamically and uniquely extracted from "trace", at least in a neighborhood of the identity matrix since the identity matrix, as a singularity of \eqref{star1}, attracts all nearby orbits, as time goes to $-\infty$. This is somewhat compatible with the classical fact that "determinant" of a matrix $B$, as an invariant polynomial, can be generated by "trace" of powers of $B$, that is $\operatorname{trace}(B^k),\;k\in \mathbb{N}$. But this dynamical interpretation we provided, needs merely trace of power $1$ not higher powers. More precisely, if we denote by $\phi$ the flow of $(*)$, then for $B$ sufficiently close to identity matrix we have $$\operatorname{Det}(B)=\exp\left(\int_{-\infty}^0 (n-\operatorname{trace})(\phi_t(B))dt\right).$$ So knowing "trace" leads us to knowing "determinant".
 A: This is a long comment, rather than an answer, so it perhaps fits here.
Another way in which the trace determines the determinant, at least locally near the identity, is via the equation $$D(e^h)=e^{T(h)}.\tag{$\dagger$}\label{dagger}$$  I have tried to prove that the OP's condition $\eqref{star3}$ implies \eqref{dagger} but, although I still think this is true, I cannot come up with a proof (any ideas?).
Regarding the existence of solutions for \eqref{dagger}, there is a clear obstruction:  if $p$ is an idempotent element, then $e^{2\pi i p}=1$, so one would need $e^{2\pi i T(p)}=1$, which is equivalent to saying that $T(p)$ is an integer.
Conversely, it is a consequence of Theorem (II.10) in my PhD thesis Rotation numbers for automorphisms of
$C^*$ algebras that the existence of projections with non-integer trace is the only possible obstruction to the existence of a determinant.
Regarding (reduced) group C*-algebras the question becomes very interesting.  If the group $G$ has torsion, then the spectral projections of any nontrivial torsion element will have non-integer trace (for the standard normalized trace, of course), so no determinant function exists.
On the other hand, if $G$ has no torsion one would hope to prove that all projections have integer trace, but doing so would solve the Kadison–Kaplansky conjecture which has been proved for many groups, including all hyperbolic groups, but I think the most general case is still open.
