Background. Exercise 36 in Rose's A Course On Group Theory reads
Prove that there is no field $F$ with $F^\times \cong F^+$.
The problem is solved in characteristic $\ne 2$ by considering $-1$ (no homomorphism $F^\times\to F^+$ is injective because $-1 \mapsto 0$) and in characteristic $2$ by considering the identity $x + x = 0$ holding for all $x$ in $F$ (no homomorphism $F^+ \to F^\times$ is injective because everything maps to $1$).
In fact, if $\operatorname{char} F = p$ (actually, these proofs only work in $F = \mathbb{F}_p$) then both $\operatorname{Hom}(F^\times, F^+)$ and $\operatorname{Hom}(F^+, F^\times)$ are trivial, because, for all $x$, $$\varphi(x) = \varphi(x^p) = p\varphi(x) = 0$$ for any homomorphism $\varphi:F^\times\to F^+$, and $$\psi(x) = \psi(x)^p = \psi(px) = \psi(0) = 1$$ for any homomorphism $\psi:F^+\to F^\times$.
Question. What can be said about $\operatorname{Hom}(F^\times, F^+)$ and $\operatorname{Hom}(F^+, F^\times)$ when $\operatorname{char}F = 0$? I can't even figure this out for $F = \mathbf{Q}$.
Update 1. $\operatorname{Hom}(\mathbf{Q}^+,\mathbf{Q}^\times) = 0$ because if $\psi:\mathbf{Q}^+\to\mathbf{Q}^\times$ is any homomorphism, then, for any $x$, $$\psi(x) = \psi\left(\frac{x}{n}+\dotsb+\frac{x}{n}\right) = \psi\left(\frac{x}{n}\right)^n\tag{1}$$ for all $n \ge 1$, so $\psi(x)$ is a square, a cube, a fourth power, etc. This is a severe restriction: $\psi(x)$ must actually be $1$. To prove this, we show that $\nu_p(\psi(x)) = 0$ for all primes $p$. This follows from (1), for then $\nu_p(\psi(x)) = n\nu_p(\psi(x/n))$, so $n \mid \nu_p(\psi(x))$ for all $n \ge 1$, and only zero has that many divisors. (Recall $\nu_p(m) = \max\{k\ge 1 : p^k\mid m\}$ for $m$ in $\mathbf{Z}$, and $\nu_p(a/b) = \nu_p(a) - \nu_p(b)$.)
Update 2. If $\operatorname{char}(F) = p$ then $\operatorname{Hom}(F^+, F^\times) = 0$ because if $\psi:F^+\to F^\times$ is a group homomorphism then $$1 = \psi(0) = \psi(px) = \psi(x)^p$$ for all $x$ in $F$, but the only solution to $X^p - 1 = (X - 1)^p$ in $F$ is $1$, so $\psi$ is constant. (Thanks Jyrki!)
The answer depends on the field. For instance, if $F=\mathbb{R}$, then the group $\mathbb{R}^+$ is a vector space over the rationals with a basis of cardinality $2^{\aleph_0}$. Since $\varepsilon\colon x\mapsto e^x$ is a group monomorphism $\mathbb{R}^+\to\mathbb{R}^\times$ and $\mathbb{R}^\times\cong\operatorname{Im}\varepsilon\times\{-1,1\}$, we have that $\mathbb{R}^\times\cong\{-1,1\}\times\mathbb{R}^+$, so there's plenty of homomorphisms $\mathbb{R}^+\to\mathbb{R}^\times$.
On the other hand, $\mathbb{Q}^\times\cong\{-1,1\}\times\mathbb{Z}^{(\mathbb{N})}$, because any nonzero rational number can be uniquely written as $$ sp_1^{k_1}p_2^{k_2}\dots p_n^{k_n} $$ where $p_1,p_2,\dots,p_n$ are distinct primes, $k_i\in\mathbb{Z}$ and $s\in\{-1,1\}$.
In particular $\mathbb{Q}^\times$ is reduced (it has no divisible subgroup), whereas $\mathbb{Q}^+$ is divisible. Thus $$ \operatorname{Hom}(\mathbb{Q}^+,\mathbb{Q}^\times) $$ is trivial, but $$ \operatorname{Hom}(\mathbb{Q}^\times,\mathbb{Q}^+)\cong \operatorname{Hom}(\{-1,1\}\times\mathbb{Z}^{(\mathbb{N})},\mathbb{Q}^+)\cong (\mathbb{Q}^+)^{\mathbb{N}} $$ is a very big group (isomorphic to $\mathbb{R}$, actually).
For $F=\mathbb{C}$ we can do similar considerations based on the fact that $\mathbb{C}^+$ is isomorphic to $\mathbb{R}^+$ and $\mathbb{C}^\times\cong (\mathbb{R}/\mathbb{Z})\times(\mathbb{R}^\times/\{-1,1\})\cong(\mathbb{R}/\mathbb{Z})\times\mathbb{R}^+$.
Thus $$ \operatorname{Hom}(\mathbb{C}^\times,\mathbb{C}^+)\cong \operatorname{Hom}(\mathbb{R}^+,\mathbb{R}^+) $$ and $$ \operatorname{Hom}(\mathbb{C}^+,\mathbb{C}^\times)\cong \operatorname{Hom}(\mathbb{R}^+,\mathbb{R}/\mathbb{Z})\times \operatorname{Hom}(\mathbb{R}^+,\mathbb{R}^+) $$ are both very big.