Does any normalized function $D$ other than determinant of matrix satisfy $D(AB) = D(A)D(B)$? Let $M(n, \mathbb{C})$ be the set of $n\times n$ matrices over $\mathbb{C}$. Let
$$
D:M(n, \mathbb{C}) \rightarrow \mathbb{C}
$$
Suppose $D$ satisfies

*

*$D(I) = 1$

*$D(AB) = D(A)D(B)$
Is it possible to prove that $D = \text{det}$ where $\text{det}$ is the usual determinant?
If not then
(1) can I see a counter-example and
(2) can we add additional, coordinate-free, conditions to $D$ to make it so?
edit: comments quickly pointed out some counter-examples including $D(A) = \text{det}(A^k)$ for positive integer $k$.
What if we add the condition
(3) $D(cA) = c^nD(A)$ for $c \in \mathbb{C}$

The comments contain a few counterexamples, including $A\mapsto \mathrm{det}(A)^k$ for $k\ge 0$. If you assume that the map $\varphi:M_n(\mathbb{C})\to \mathbb{C}$ is continuous in the entries of the matrix, every possibility will be of the form $f\circ \mathrm{det}$, where $f$ is a group homomorphism $\mathbb{C}^* \to \mathbb{C}^*$, extended continuously to $\mathbb{C}$.
To see this, observe that $\mathrm{GL}_n(\mathbb{C})$ is dense in $M_n(\mathbb{C})$, so determining the map $\varphi$ amounts to determining it on $\mathrm{GL}_n(\mathbb{C})$. Now $\psi = \varphi_{|\mathrm{GL}_n(\mathbb{C})}$ is a group homomorphism to $\mathbb{C}^*$. Since the latter is abelian and the derived subgroup of $\mathrm{GL}_n(\mathbb{C})$ is $\mathrm{SL}_n(\mathbb{C})$, the map $\psi$ factors through $\mathrm{det}:\mathrm{GL}_n(\mathbb{C})\to\mathrm{GL}_n(\mathbb{C})/\mathrm{SL}_n(\mathbb{C})=\mathbb{C}^*$.
Note that there are many group automorphisms $f:\mathbb{C}^*\to\mathbb{C}^*$, including $x\mapsto x^k$ and $x\mapsto \overline{x}$, among many others (in a sense, most cannot be written down).
