Automorphisms of U(1) This probably a very simple question but I would appreciate some clarification. 
So $U(1)$ is an Abelian Lie group. As it is Abelian, all elements of the group commute with each other, this means the centre of the group is the group itself.
Since 
     $$\text{Inn}(G)=G/Z\implies \text{Inn}(U(1))=\mathcal{I} $$
I would like to know the outer automorphisms
$$\text{Out}(G)=\text{Aut}(G)/\text{Inn}(G)  $$
However I do not know the automorphism group in order to make this deduction.
 A: First, we need to distinguish between the continuous automorphism group and the abstract automorphism group.
I claim that there is a unique continuous non-identity automorphism of $S^1$.
First, since $S^1$ is abelian, the inversion map $i:S^1\rightarrow S^1$ is a continuous isomorphism.
Why is it the only one?  Let $g:S^1\rightarrow S^1$ be any continuous automorphism.  If $\pi:\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}\cong S^1$ is the canonical projection, then the map $\pi \circ g$ lifts to a unique continuous map $\tilde{g}:\mathbb{R}\rightarrow \mathbb{R}$ with $\tilde{g}(0) = 0$.  One can easily verify that $\tilde{g}$ is a homomorphism, that it $\tilde{g}$ satisfies the Cauchy functional equation.  In particular, $\tilde{g}(x) = cx$ for some constant $c\in \mathbb{R}$.
Now, because $\tilde{g}(\mathbb{Z})\subseteq \mathbb{Z}$, it follows that $g(1) = c\in \mathbb{Z}$.  This establishes that every continuous homomorphism $S^1\rightarrow S^1$ is of the form $z\rightarrow z^c$ with $c$ an integer.  (Here, we are thinking of $S^1$ as the unit complex numbers).  Then note that if $|c|\neq 1$, this map is not injective.  So there are precisely two continuous automorphisms of $S^1$, the identity and the inverse.
Now, if one is simply interested in the group theoretical automorphism group of $S^1$, then there are a ton more (assuming you believe in the axiom of choice.)
To see this, note that, as shown in this surprising post, $S^1$ is, group theoretically, isomorphic to $\mathbb{C}^\times$.  In particular, every automorphism of $\mathbb{C}$ as a field can be thought of as a group isomorphism of $S^1$.  According to this post, there are at precisely $2^{|\mathbb{R}|}$ such automorphisms.
(Note that it's possible that an automorphism of $S^1\cong \mathbb{C}^\times$ does not extend to a field isomorphism of $\mathbb{C}$, so there may be even more automorphisms of $S^1$.  This may also allow one to create discontinuous automorphisms even in the absence of choice, but I'm not sure what happens in that case.)
A: Write $U(1)=\{e^{ix},x\in R\}$. Let $f$ be an automorphism of $U(1)$, $f(e^{ix})=e^{ig(x)}$, 
Suppose $f$ is continue, $f(e^{ix}e^{iy})=f(e^{i(x+y)})=$ $f(e^{ix})f(e^{iy})$ $=e^{ig(x+y)}=e^{ig(x)+g(y)}$, implies that $h(x,y)=e^{i(g(x+y)-g(x)-g(y))}=1$. Since $h$ is continue, we deduce that there exists such that $g(x+y)-g(x)-g(y)=2\pi n$.
Write $l(x)=g(x)+2\pi n$,you deduce that $l$ is linear, $l(x)=ax$ and $f(x)=e^{iax}$.
