TITLE: Isomorphism of sheaves
QUESTION [10 upvotes]: Let $X$ be a scheme and $\mathscr{F}$, $\mathscr{G}$ sheaves on $X$. Suppose we know that for every $x\in X$ the stalks are isomorphic:
$$ \mathscr{F}_x \cong \mathscr{G}_x $$
Can we conclude immediately from here that $\mathscr{F}\cong \mathscr{G}$, or do we first need to find a morphism of sheaves $f:\mathscr{F}\to \mathscr{G}$ such that $f_x$ induces the above isomorphisms on the stalks for every $x$ ?
Remark: Proposition 1.1 (page 63) of Hartshorne says that: 

A morphism of sheaves $f : \mathscr{F} \to \mathscr{G}$ on a
  topological space $X$ is an isomorphism iff the induced map on stalks
  $f_x : \mathscr{F}_x \to \mathscr{G}_x$ is an isomorphism for every
  $x\in X$.

Extra question:
What can we say, instead, if we have isomorphisms between every fiber of $\mathscr{F}$ and $\mathscr{G}$ ?

REPLY [7 votes]: If this were indeed true then all line bundles on any scheme would be isomorphic! However this is not true as for example take $\Bbb{P}^1$ that has an integers worth of non-isomorphic line bundles. The point is that an isomorphism $\varphi_x$ of $\mathcal{O}_{X,x}$-modules need not glue to give an isomorphism of $\mathcal{O}_X$-modules (for example how is it even clear to go from an isomorphism of modules at every point to an isomorphism of sheaves globally?).
However if say given a priori a morphism of sheaves $\varphi : \mathcal{F} \to \mathcal{G}$  that is an isomorphism on the stalks at every point, then it is indeed true that $\varphi$ will be an isomorphism. As for your last question, it is not clear to me what you mean by the fiber of a sheaf.