$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\id}{\mathrm{id}}$I like your definition of an antimorphism and I'll raise you one: if $\cat{C}$ comes with an autoequivalence $F$, an $F$-antimorphism 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 antimorphism inverse (henceforth known as an "antigroup"), 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 an antimorphism 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, 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 an $F$-antigroup in $\cat{C}$, but check their properties in $\cat{D}$ (since $S$ is faithful, this won't result in any errors).

So, say that an $F$-antigroup in $\cat{C}$ is an object $G$ 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.

$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\id}{\mathrm{id}}$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.

$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\id}{\mathrm{id}}$I like your definition of an anti-inverse and I'll raise you one: if $\cat{C}$ comes with an autoequivalence $F$, an $F$-antimorphism 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 antimorphism inverse (henceforth known as an "antigroup"), 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 an antimorphism 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, 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 an $F$-antigroup in $\cat{C}$, but check their properties in $\cat{D}$ (since $S$ is faithful, this won't result in any errors).

So, say that an $F$-antigroup in $\cat{C}$ is an object $G$ 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.

$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\id}{\mathrm{id}}$I like your definition of an antimorphism and I'll raise you one: if $\cat{C}$ comes with an autoequivalence $F$, an $F$-antimorphism 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 antimorphism inverse (henceforth known as an "antigroup"), 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 an antimorphism 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, 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 an $F$-antigroup in $\cat{C}$, but check their properties in $\cat{D}$ (since $S$ is faithful, this won't result in any errors).

So, say that an $F$-antigroup in $\cat{C}$ is an object $G$ 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.