Confusion about how the first cohomology classifies torsors This question is inspired by, but is independent of: Sheaf Description of G-Bundles
Line bundles are classified by $H^1(X,\mathcal{O}^\times_X)$. We also know that in general that $H^1(X,G)$, where $G$ is a sheaf from open sets in $X$ to $Grps$, classifies $G$-torsors over X.
With this insight in mind: $\mathcal{O}^\times_X$-torsors should correspond to line bundles. Indeed, if $P$ is one, then $\mathcal{O} _X \times _{\mathcal{O} _X^\times}P$ gives the desired line bundle, and all line bundles are achieved this way (see the question I linked to).
My question is about the more mundane $H^1(X,\mathbb{C})$, which can be thought of as $H^1(X,\underline{\mathbb{C}})$ where $\underline{\mathbb{C}}$ is the constant sheaf $\mathbb{C}$ (which I think of as a sheaf going to $Grps$). These should supposedly correspond to $\mathbb{C}$-bundles over $X$. Which appears to be line bundles. But of course there's no reason for $H^1(X,\mathbb{C})$ to equal $H^1(X,\mathcal{O}^\times_X)$... My intuition is that this should correspond to the more naive version of fiber bundles that doesn't involve a structure group. Do you have any thoughts?
 A: I think the right thing to look at is $H^1(X, \mathbb C^\*)$. This classifies line bundles with  a flat connection, or equivalently, line bundles with locally constant transition functions.
Now the natural embedding $\mathbb C^\* \to \mathbb O_X^\*$ induces on a map on cohomology $H^1(X, \mathbb C^\*) \to H^1(X, \mathbb O_X^\*)$ which is forgetting the flat connection.
A: The general principle is: if you have some objects which are locally trivial but globally possibly not trivial then the isomorphism classes of such objects are classified by $H^1(X,\underline{Aut})$, where $\underline{Aut}$ is the sheaf of automorphisms of your objects.
So, if your objects locally are $U\times \mathbb A^n$ (i.e. vector bundles) or they are $\mathcal O_U^{\oplus n}$ (i.e. locally free sheaf) then either of these are classifed by $H^1(X, GL(n,\mathcal O))$. For $n=1$, you get $GL(1,\mathcal O)=\mathcal O^*$. 
Now what is $\mathbb C$ an automorphism group of? Certainly not of line bundles (zero has to go to zero).
A: Your confusion seems to stem from the difference between topological bundles and algebraic/holomorphic bundles.
For a scheme $(X, \mathcal{O}_X),$ you say that $H^1(X, \mathcal{O}_X^{\times})$ classifies line bundles on $X$.  This is true, as long as you mean algebraic line bundles.  If, for example, $X$ is something like a complex algebraic variety with the analytic topology (which seems to be how you're thinking about it, perhaps), then there can certainly be plenty of topological line bundles on $X$ which aren't algebraic.
Also, as Lucas Culler notes in the comments to your question, a line bundle is not a torsor under the additive group $\mathbb{C}$.  Instead, it is actually a $\mathbb{C}^{\times}$ torsor, if you remove the zero section.  Algebraically, a torsor for the additive group $\mathbb{C}$ (which, by the way, is often denoted $\mathbb{G}_a$ to avoid confusion) is classified by all extensions of $\mathcal{O}_X$ by $\mathcal{O}_X$.
Anyway, it is true that (maybe with some mild assumptions on your space $X$) that $H^1(X,G)$ classifies topological $G$-bundles on $X$.
