Anti-isomorphisms
These questions relate to the following construction:
Let $(G,\cdot)$ be a group. Define its opposite $G^{op}$ to be the group $(G,*)$ where we define $$x*y=y\cdot x.$$
One may check that this indeed forms a group. Then, an antihomomorphism $f:G\rightarrow H$ is exactly a homomorphisms $G\rightarrow H^{op}$ or, equivalently, a homomorphism $G^{op}\rightarrow H$.$^1$ Essentially all of your results follow from the fact that antihomomorphisms are homomorphisms - their domain is just the opposite of what its marked as.
In some sense, an anti-isomorphism is the same as a normal isomorphism. This is because $G$ and $G^{op}$ are always isomorphic - the map $f:G\rightarrow G^{op}$ defined by $f(g)=g^{-1}$ is an isomorphism.$^2$ Thus, if $G$ and $H$ are anti-isomorphic, they are actually isomorphic (and vice versa).
$^1$ One might observe that a homomorphism $G^{op}\rightarrow H^{op}$ is also a homomorphism $G\rightarrow H$ and vice versa.
$^2$ This follows easily since $f(g\cdot h)=(g\cdot h)^{-1}=h^{-1}\cdot g^{-1}=f(h)\cdot f(g)=f(g)*f(h)$. Obviously, this is bijective since inversion is an involution.
An antihomomorphism can be used, for example, to turn a left group action into a right group action.
It is dual to the notion of homomorphism: if we see groups as categories with one object, then homomorphisms are covariant functors, antihomomorphisms are contravariant functors. Therefore, it is very natural that the two notions are very much similar.
Sometimes anti-isomorphisms arise naturally: the map $g\to g^t$ on $GL_n(\mathbb{R})$ or $g\to g^\dagger$ on $GL_n(\mathbb{C})$, for example. You can interpret an anti-homomorphism $G \to H$ as a homomorphism $G \to H^{\text{op}}$, though, where $H^{\text{op}}$ is the group with the same underlying set as $H$ and group operation $x.y = yx$. The subject is not particularly interesting, though, since the map $H \to H^{\text{op}}$ given by $x \to x^{-1}$ is an isomorphism. For (noncommutative) rings, though, the situation is more interesting: the rings $A$ and $A^{\text{op}}$ are not necessarily isomorphic.