This is a follow up question on: Product not definable by addition in Th($\mathbb{R},+,\cdot,0,1,<)$.
I've shown that the only automorphism of $(\mathbb{R}, 0,1, +, - , \cdot)$ is the identity map.
But I'm having trouble to find an example non-trivial automorphism of $(\mathbb{R},0, 1, +)$ that fixes $\mathbb{R}$ pointwise and show it's it's not definable.
Thanks in advance!
You seem to have a bit of confusion about the relationship between definable subsets and automorphisms. The result we want to use here is that, if $A\subseteq M$ is a subset of an $\mathcal{L}$-structure $\mathfrak{M}$, then every subset of $M^n$ that is definable over $A$ (ie, that is definable by a formula in the expanded language $\mathcal{L}\cup A$) is fixed by any automorphism of $\mathfrak{M}$ that fixes $A$ pointwise. Can you see why this is?
In particular, if we're trying to show that multiplication is not definable in $(\mathbb{R}, 0, 1, +)$, (note that "definable" is just shorthand for "definable over $\emptyset$"), it suffices to find any automorphism of $(\mathbb{R}, 0, 1, +)$ that does not preserve multiplication. (Because any such map vacuously fixes $\emptyset$ pointwise.) In particular, we're not looking for automorphisms of $\mathbb{R}$ that "fix $\mathbb{R}$ pointwise", for as Alex points out any such map is the identity map by definition.
Now, if you know that $(\mathbb{R}, 0, 1, +, \cdot)$ has no non-trivial automorphisms, then we just need to find any non-trivial automorphism of $(\mathbb{R}, 0, 1, +)$, for any such map will not preserve multiplication. To do this, extend the element $x_0=1$ to a basis $(x_i)_{i\in I}$ of $\mathbb{R}$ considered as a $\mathbb{Q}$-vector space. Then any non-trivial permutation of the basis defines a non-trivial linear isomorphisms from $\mathbb{R}$ to itself; choose any such isomorphism that fixes $x_0=1$. By definition of linear maps, this map will fix $0$ and preserve addition, and since it also preserves $1$ by construction it is a non-trivial $\mathcal{L}$-automorphism of $\mathbb{R}$, as desired.