Is addition on $\mathbb{R}$ unique up to automorphism? Consider the set of real numbers $\mathbb{R}$, which we know is the only complete ordered field up to isomorphism. 
If we fix the following axioms for the 'addition' operation $\bot:\mathbb{R}^2\to \mathbb{R}$: 


*

*Associativity

*Commutativity

*Identity element 0 (not that there exists some 0 element, but that the identity element is actually the canonical 0).

*All elements have their own unique inverse


[this defines an abelian group over the set $\mathbb{R}$ with identity 0]
and finally


*$\bot$ is distributive over multiplication: $\forall a,b,\lambda\in \mathbb{R}, $ $\lambda\cdot (a\bot b)=(\lambda \cdot a) \bot (\lambda \cdot b).$



Do these axioms uniquely determine addition on $\mathbb{R}$ up to automorphism? 

The example of $a\bot b := (a^3+b^3)^{1/3}$ shows that these axioms do not uniquely determine addition. However, the automorphism $f(x)=x^3$ makes this operation still effectively the same as addition. The question is whether all operations $\bot$ which adhere to these axioms have an automorphism with addition. 
And as a secondary question, if it turns out that they do:

Is the fifth axiom necessary, or is the only abelian group with identity 0 on the real numbers addition (up to automorphism)?

EDIT: The second part has been answered, and the answer is yes, it is necessary.
 A: Your axioms do not determine addition uniquely.  Note that if $K$ is any field and $f:\mathbb{R}^\times\to K^\times$ is an isomorphism of the multiplicative groups, then the operation $(x,y)\mapsto f^{-1}(f(x)+f(y))$ (and $(x,0)\mapsto x$, $(0,y)\mapsto y$) will satisfy your axioms, and will only be equivalent to ordinary addition in your sense if $K$ is isomorphic to $\mathbb{R}$ as a field.
Now note that the abelian group $\mathbb{R}^\times$ is not too hard to understand.  It is the direct sum of the subgroups $\{\pm1\}$ and $\mathbb{R}_+$, and $\mathbb{R}_+$ is a vector space over $\mathbb{Q}$ since every positive real number has an $n$th root for any $n$.  More generally, the same description holds for $K^\times$ if $K$ is any ordered field in which every positive element has an $n$th root for any $n$.  Since two $\mathbb{Q}$-vector spaces of the same uncountable cardinality are isomorphic, it follows that if $K$ is any ordered field of the same cardinality as $\mathbb{R}$ in which every positive element has an $n$th root for any $n$, then $K^\times\cong\mathbb{R}^\times$.
However, such a field need not be isomorphic to $\mathbb{R}$, and so can give rise to a different "addition" on $\mathbb{R}$ satisfying your axioms.  For instance, you could take $K$ to be any non-archimedean real-closed field of cardinality $2^{\aleph_0}$ (e.g., the real closure of $\mathbb{R}(x)$ where $x$ is infinitely large).
