Call two functors two functors $H,H':S\longrightarrow T$ *weakly equivalent*, or *equivalent up to a self-equivalence of the source category*, iff
there exists a self-equivalence of $s:S \longrightarrow S$ such that functors $H$ and $H'\circ s$ are equivalent.

Can this property "weakly equivalent" be nicely reformulated in the language of higher category theory? or maybe homotopy theory?

Are there theorems claiming that any two functors with certain properties (not involving choice) are weakly equivalent?

That is, for such a theorem to be interesting, the properties should not explicitly involve an arbitrary choice, e.g. it should not say: choose a topology, bijection, self-equivalence and then construct the functor in the following way. Rather, a functor should be described in terms of preserving some structure etc.

Below is the original question which was phrased very confusingly, it seems. I hope now it maybe is clearer.

Say a functor is "well-defined up to a self-equivalence of the source category" by certain properties/definition/construction iff, well, for any two functors $H,H':S\longrightarrow T$ with satisfying these properties/definition/obtained by this construction, there exists a self-equivalence of $s:S \longrightarrow S$ such that functors $H$ and $H'\circ s$ are equivalent.

Is there a nice way to reformulate this property "a functor unique up to self-equivalence of the source category", say in the language of 2-categories?

Are there any interesting examples of properties/definitions/constructions NOT involving arbitrary choice and yet such that the functor is well-defined up to a self-equivalence of source category ?

I am mostly interested to see an "algebraic" definition of a functor between "algebraic" categories which is well-defined up to self-equivalence but not well-defined.

nota functor. $\endgroup$ – Fernando Muro Aug 17 '12 at 8:57chosenproducts, then "the product functor" $C \times C \to C$, while not specified, exists and is "defined well enough" since any two choices are canonically isomorphic. A property P of functors would be "well-defining" if any two functors satisfying P are isomorphic, e.g., a universal property P. (But the question needs work.) $\endgroup$ – Todd Trimble♦ Aug 17 '12 at 10:54someproperty. Given a functor $F: C \to D$, define property P on functors $G: C \to D$ by the condition that $G = F$ identically. Any two functors satisfying property P are canonically isomorphic. This silly example indicates that you probably need to sharpen what you really intend by "well-defined". $\endgroup$ – Todd Trimble♦ Aug 17 '12 at 15:45