Interpretation of alternative group structures on a given group Let $(G,e,\circ) $ be a group with $e$ the identity element and $a \in G$ and $\circ$ the group operation. Then we can form a new group $(G_a,a,\circ_a)$ with the same underlying set as $G$ and $x \circ_a y = x\circ a^{-1}\circ y$. 
It is easy to verify that $a$ is the identity element for $G_a$ with inverse of $x$ being $a\circ x^{-1}\circ a$, $x^{-1}$ being the inverse in $G$. 
Edit: There are isomorphisms from $G$ to $G_a$ given by $x \mapsto a \circ x$ and $x \mapsto x \circ a$.
This seems to work out quite naturally and corresponds to something intuitive(shifting the origin in $(\Bbb R^n,+,0)$). However, I have never seen it used in any context. Is this indicative of something deeper or is it some trivial insight that I am blowing up? 
PS: I am not sure of the optimal tags for this question. Please feel free to retag if you see fit.
 A: There is a little bit going on here, and the intuition from "shifting the origin" in the real numbers is exactly right. In general, there is the notion of a group $G$ acting on a set $X$, where given $g \in G$ and $x \in X$, we can form $g \circ x$. We can think of a group as acting on itself by multiplication. A group acting on itself in this way acts freely and transitively in that, for $x,y \in X$, there is a unique $g \in G$ with $g \circ x = y$ (transitivity means existence, freeness means uniqueness).
In general, whenever a group $G$ acts on a set $X$ freely and transitively, the set $X$ is essentially just a copy of the group. However, it's an affine copy in that we "forgot" which element was the identity. That is, we can't actually multiply elements of $X$ by each other, but instead we get a ternary operation that sends $(x,y,z)$ to $xy^{-1}z$: it looks at the unique map sending $y$ to $x$, and applies that map to $z$. We can recover the group if we pick any element $a \in X$ to serve as the identity: build a map $f_a : G \to X$ by saying $f_a(g) = g \circ a$. This is a bijection exactly because the action is free and transitive. By putting $X$ in bijection with the group $G$, that gives $X$ the structure of a group. With that group structure, $a$ is the identity element, and $f_a$ is a group isomorphism.
The explicit construction you give is essentially using the fact that a group acts freely and transitively on itself by multiplication to build an affine copy of $G$, and then turning that affine copy into a group (isomorphic to the original) by picking an identity element from the affine copy.
I leave finding the correct axioms for saying when a set with a ternary operation is an "affine copy of a group" as an exercise.
