Is $e^x$ the only morphism for addition and multiplication? I find it interesting that the only (more or less) function that is equal to its own derivative also happens to be a morphism from the reals under addition to the positive reals under multiplication. This would be even more interesting if it were the only such morphism.
Is this the case?
 A: Once you know that $\def\R{\mathbb{R}}(\R_{>0},\cdot)$ is isomorphic to $(\R,+)$ via $\exp$, what you're asking is classifying the additive isomorphisms of $(\R,+)$ onto itself.
If the isomorphism is supposed to be continuous, then it is of the form $x\mapsto ax$, for some $a\in\R$, $a\ne0$. But there are many others, because $\R$ is a vector space over $\def\Q{\mathbb{Q}}\Q$ and vector space isomorphisms are in particular additive isomorphisms of $(\R,+)$ onto itself.
Since the cardinality of a basis of $\R$ over $\Q$ is the same as $|\R|$, there are at least $|\R|$ permutations of the basis, which give as many isomorphisms.
A: The exponential function and the trivial morphism are the only group homomorphisms $(ℝ,+,0) → (ℝ_{>0},·,1)$ which are differentiable and equal to their derivatives.
Any differentiable function $f \colon ℝ → ℝ$ such that $f' = f$, observing $\tfrac{f}{\exp}$ is constant by differentiating, can be written as $f = f(0)·\exp$. For $f$ to be a group homomorphism $(ℝ,+,0) → (ℝ_{>0},·,1)$ you need $f(0) = 1$, so then $f = \exp$.
Edit: Okay, I just realized that you wasn’t asking for that and that I misread your question, but here continues the more interesting part for you:
There are non-continuous group homomorphisms $(ℝ,+,0) → (ℝ_{>0},·,1)$, since any $ℚ$-base of $ℝ$ defines a $ℚ$-automorphism of $ℝ$, which also gives an underlying group homomorphism $(ℝ,+,0) → (ℝ,+,0)$. Postcomposing $\exp$ yields many non-differentiable group homomorphisms.
