Given a topological field $K$ that admits a non-trivial continuous exponential function $E$, must every non-trivial continuous exponential function $E'$ on $K$ be of the form $E'(x)=E(r\sigma (x))$ for some $r \in K$* and $\sigma \in Aut(K/\mathbb{Q})$?
If not, for which fields other than $\mathbb{R}$ is this condition met?
Thanks to Zev
On
Here's a $p$-adic example to consider. In ${\mathbf C}_p$ (the $p$-adic complex numbers) let $$D = \{x \in {\mathbf C}_p : |x|_p < (1/p)^{1/(p-1)}\},$$ which is the disc of convergence of the $p$-adic exponential series $\exp(x) = \sum_{n \geq 0} x^n/n!$. Then $\exp(D) = 1 + D$, and $1+D$ is a multiplicative subgroup of $1 + {\mathfrak m}_p$, where ${\mathfrak m}_p = \{x \in {\mathbf C}_p : |x|_p < 1\}$. The group $1 + {\mathfrak m}_p$ is divisible, so for any abelian group $G$ and subgroup $H$, any group homomorphism $f \colon H \rightarrow 1 + {\mathfrak m}_p$ extends to a group homomorphism $G \rightarrow 1 + {\mathfrak m}_p$. This extension from $H$ to $G$ is based on Zorn's lemma, and if you look at the proof you see that it involves gazillion choices along the way (if $[G:H] = \infty$). So there's huge amount of flexibility in making the extension.
We will apply this to the additive group ${\mathbf C}_p$ and subgroup $D$: the homomorphism $\exp \colon D \rightarrow 1 + {\mathfrak m}_p$ extends to a homomorphism ${\mathbf C}_p \rightarrow 1 + {\mathfrak m}_p$ in many different ways. Since this extended homomorphism is the power series $\exp(x)$ on a neighborhood of $0$, the continuity of $\exp$ at $0$ implies the continuity of any extension to a homomorphism on ${\mathbf C}_p$. I haven't thought carefully about why you can't write all such extensions in terms of one of them using premultiplication, but it feels clear to me from the Zorn's lemma basis for making the extension that such a relation among all such extensions is not true.
It seems that as-stated, the answer is false. I'm not satisfied with the following counterexample, however, and I'll explain afterwards.
Take $K = \mathbb{C}$ and let $E(z) = e^z$ be the standard complex exponential. Take $E'(z) = \overline{e^z} = e^{\overline{z}}$, where $\overline{z}$ is the complex conjugate of $z$. Then $E'(z)$ is not of the form $E(r z)$, and yet is a perfectly fine homomorphism from the additive to the multiplicative groups of $\mathbb{C}$.
Here's why I'm not satisfied: you can take any automorphism of a field and cook up new exponentials by post-composition or pre-composition. In the case I mentioned, these two coincide.
This won't work in $\mathbb{R}$ because there are no nontrivial continuous automorphisms there. I would be interested in seeing an answer to a reformulation to this problem that reflected this.