Characterizing continuous exponential functions for a topological field 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

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.
