So choose $\iota \colon \text{U}(1) \to \text{U}(n); z \mapsto z \cdot E_n$ to be your usual embedding. This is of course a Lie-group homomorphism and I want to know if there is a Lie-group homomorphism $\phi \colon \text{U}(n) \to \text{U}(1)$ which satisfies $\phi(z \cdot A) = z \cdot \phi(A)$ for each $z \in \text{U}(1)$ and $A \in \text{U}(n)$.
The Determinant only satisfies $\text{det}(z \cdot A) = z^n \cdot \text{det}(A)$. It would be sufficient to have a Lie-group homomorphism $\psi \colon \text{U}(n) \to \text{U}(n-1)$ such that $\psi(z \cdot A) = z \cdot \psi(A)$ for we could then iterate it.
I'm not sure if such a map can exist. It would imply that the $\text{U}(1)$-principal bundle $t \colon \text{U}(n) \to \mathbb{P}\text{U}(n)$ given by the usual projection would then have a smooth section $\sigma \colon \mathbb{P}\text{U}(n) \to \text{U}(n)$ given by $\sigma([A])= \phi(A)^{-1} \cdot A$. Hence this would imply that $\text{U}(n) \cong \mathbb{P}\text{U}(n) \times \text{U}(1)$ as $\text{U}(1)$-principal bundle.
Such a homomorphism cannot exist for $n>1$. To see this, look the the Lie algebras in question. The algebra $\mathfrak{u}(1)$ is $i\mathbb R$ while $\mathfrak{u}(n)\cong i\mathbb R\oplus\mathfrak{su}(n)$ and for $n>1$, $\mathfrak{su}(n)$ is simple. The summand $i\mathbb R$ corresponds to multiples of the identity, the second summand to matrices with vanishing complex trace. Now a Lie algebra homomorphism $\mathfrak{u}(n)\to\mathfrak{u}(1)$ has to vanish on $\mathfrak{su}(n)$ (otherwise its kernel would be a non-trivial ideal). So the only such homomorphisms are $X\mapsto a\cdot tr(X)$ for a real number $a$. Now for the homomorphism you are looking for, the derivative by definition would have to be $X\mapsto \frac1n\cdot tr(X)$. But this Lie algebra homomorphism does not integrate to the group $U(n)$ since $\exp(2\pi i\mathbb I)=\mathbb I$ whereas $\exp(\frac{2\pi i}{n})\neq 1$ for $n>1$. (So this corresponds to the fact that there is no $k$th root as a homomorphism $U(1)\to U(1)$.)
The argument via simplicity of $\mathfrak{su}(n)$ also shows that there are no interesting homomorphisms $U(n)\to U(n-1)$.