No, not necessarily, since a transformation involves "higher structure" in addition to its 1-cell components. For example, let C be the "walking arrow" category with two objects 0 and 1 and one nonidentity morphism from 0 to 1, and let D be an abelian group regarded as a (2,1)-category with one object and one morphism, and thereby as an (∞,1)-category. There is exactly one functor from C to D. An endo-natural-transformation of this functor consists of giving, for each object of C, a morphism in D, and for each morphism in C, a 2-cell in D, subject to certain axioms.

There is only one morphism in D, so any two such transformations will have equal 1-cell components, but their 2-cells might be different. The axioms say that the 2-cell corresponding to an identity morphism is the identity, and that the 2-cell corresponding to a composite is the composite of the 2-cells corresponding to the individual factors; thus for the C and D above, to give a transformation is exactly to give an arbitrary element of the abelian group D (the 2-cell component corresponding to the single nonidentity arrow of C).

Now two transformations are equivalent when there is a modification between them, which consists of for each object of C, a 2-cell in D between the corresponding 1-cell components, and for each morphism of C, a 3-cell relating the 2-cell components etc. Since 3-cells in D are all identities, this latter means that there is a diagram that must commute. Since C has one object, a modification thus consists of giving a single element of the abelian group D, and the diagram which must commute means that it *conjugates* the element corresponding to the first transformation to the element corresponding to the second one. Thus, the equivalence classes of such transformations (all of which have the same 1-cell component) are the conjugacy classes of D.