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This answer is in the context of sheaves of sets. If you meant sheaves of groups, please say so.

If you look at the 'espace etale' (wikipedia) of a sheaf $\mathcal{O}(X)^{op} \mathrm{Open}(X)^{op} \to Set$ (which is defined by taking the union of stalks and topologising appropriately - fibres of the canonical map to $X$ are then discrete spaces), then continuous maps of these over $X$ correspond to sheaf maps. The maps of stalks give a map of the underlying sets. An obvious sufficient condition that the collection of maps of stalks gives a map of sheaves is that this map is continuous.

(If you actually have a sheaf of groups, then you get a group object over $X$. All you need to check is the continuity - the homomorphisms of stalks give a homomorphism of the underlying set of the group object)

show/hide this revision's text 1

This answer is in the context of sheaves of sets. If you meant sheaves of groups, please say so.

If you look at the 'espace etale' (wikipedia) of a sheaf $\mathcal{O}(X)^{op} \to Set$ (which is defined by taking the union of stalks and topologising appropriately - fibres of the canonical map to $X$ are then discrete spaces), then continuous maps of these over $X$ correspond to sheaf maps. The maps of stalks give a map of the underlying sets. An obvious sufficient condition that the collection of maps of stalks gives a map of sheaves is that this map is continuous.

(If you actually have a sheaf of groups, then you get a group object over $X$. All you need to check is the continuity - the homomorphisms of stalks give a homomorphism of the underlying set of the group object)