$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\id}{\mathrm{id}}$I like your definition of an antimorphism (which, following Ben Webster's suggestion, I will call a "heteromorphism") and I'll raise you one: if $\cat{C}$ comes with an autoequivalence $F$, an $F$-heteromorphism is by definition a morphism $X \to F(X)$. Why work in this generality? Here's why.
I see one big issue with defining a group with heteromorphism inverse, namely, how one is to state the inversion property:
$$G \xrightarrow{\Delta} G \times G \xrightarrow{i \times \id} G \times G \xrightarrow{m} G$$
if in fact we have $i \colon G \to F(G)$; how can we get both of the latter factors the same so that $m$ may be applied? My answer is, philosophically: since a heteromorphism is a morphism after allowing the loss of some structure, we must check this diagram also after forgetting that structure. This will turn $F$ into the identity.
Here's what I mean. Let $S \colon \cat{C} \to \cat{D}$ be some "structure-forgetting" faithful functor that preserves products, and suppose $F \colon \cat{C} \to \cat{C}$ acts fiberwise for $S$, in that $SF = S$ (I suppose more generally we could also specify $\phi \colon SF \cong S$). For example, $S$ could be "forgetting the Poisson structure" or "forgetting the multiplication in a noncommutative ring" or "forget the directions of arrows in categories". Correspondingly, $F$ would be "take the negative Poisson structure" or "take the opposite ring" or "take the opposite category". We will define all the morphisms for a group object in $\cat{C}$, but check their properties in $\cat{D}$ (since $S$ is faithful, this won't result in any errors).
So, say that a $(\cat{C},F)$-group object (any suggestions for a better name for this?) is an object $G \in \cat{C}$ together with morphisms
$$m \colon G \times G \to G, \qquad i \colon G \to F(G), \qquad u \colon 1 \to G$$
constituting a group object structure on $S(G)$. In particular, the above diagram reads
$$S(G) \xrightarrow{\Delta} S(G) \times S(G) \xrightarrow{S(i) \times \id} SF(G) \times S(G) \xrightarrow{m} S(G),$$
where we have $SF = S$ by definition.
Note that the inverse $i$, if it exists, is unique, since $S(i)$ is unique in $\cat{D}$ and $S$ is faithful. So it is, as for regular group objects, not a structure but a property. Note also that I omit the notation $S$ from "$(\cat{C},F)$-group object" because it is enough that some $S$ exist and that $F$ act on its fibers, but it doesn't matter which one we use.
Deligne does not do what you seem to want, which is give a theory internal to super vector spaces. To do so is probably an open question. The obvious version of Hopf algebras in SVect does not work: the category of comodules admits an action by the lines in SVect (commuting with the fiber functor), and thus cannot recover categories like ordinary vector spaces.
Deligne's theory is external, mixing Vect and SVect. A supergroup is an algebraic group internal to the algebraic geometry of super vector spaces (that is, a cocommutative Hopf algebra in that symmetric monoidal category, subject to some conditions). An "external group" is a supergroup with a distinguished element (of $G(k)$) of order at most 2. We do not look at all representations of the group, but only ones on which the distinguished element acts as parity (and we only allow distinguished elements such that the adjoint representation is allowed). Deligne's theorem is that every nice category is this subcategory of representations of an external group. We recover the group as automorphisms of the fiber functor to SVect. As you note, the automorphisms of the identity functor of SVect is $Z/2$; that is its external group, with the nontrivial element distinguished. Vect is representations of the trivial group. These are the initial and final external groups, and they are not isomorphic, so external groups are not group objects in some category. In general, internal representations of a supergroup are representations of an external group which is the semidirect product of the two element group acting on the original supergroup. Representations of ordinary algebraic groups reuse the group as the external group, with the trivial element distinguished.
A fairly easy example is that graded vector spaces with the signed symmetry are representations the external group $G_m$, with $-1$ distinguished. Your calculations of the automorphisms of the fiber functor on chain complexes and of the full category of representations of $G=G_m \ltimes G^{0|1}_a$ each imply that the relevant subcategory of its representations are chain complexes of ordinary vector spaces.
This is all based on just the first few pages of Deligne's paper. I don't think he has a name for what I call "external groups"; the Mueger appendix you cite calls them supergroups, but I think that's a bad choice.
Best Answer
The following result holds.
See A. V. Mikhalev, G. Pilz: The Concise Handbook of Algebra, p.74 and the references given therein.