Is there a commutative operation for which the inverse of the operation is also commutative? For instance addition is commutative, but the inverse, subtraction, is not.
$$
5+2 = 2+5\\
5-2 \neq 2-5
$$
Same for multiplication/division:
$$
5\times4 = 4\times5\\
5/4 \neq 4/5
$$
So is there a group operation $\circ$ with the inverse $\circ^{-1}$ such that
$$
a\circ b = b\circ a\\
a\circ^{-1}b = b\circ^{-1}a
$$
 A: If $G$ is an abelian group, then $ab^{-1}=ba^{-1}$ for all $a,b\in G$ if and only if each element of $G$ has order at most $2$.
A: The notion of inverse of an element can still be defined in the setting of inverse semigroups, which generalize groups. 
An inverse semigroup is a semigroup $S$ such that, for every $x \in S$, there is a unique $x^*$ such that $x x^* x = x$ and $x^* x x^* = x^*$. 
Then it can be seen that the operation $/$ defined by $x / y = x y^*$ is commutative if and only if $x = x^3$ for all $x$, if and only if $x = x^*$ for all $x$. 
In particular, in every semilattice (a semilattice being a special case of inverse semigroup) the operation $/$ is commutative. 
Edit. One can also put the result the following way. The operation $/$ is commutative if and only if the multiplication on $S$ is commutative and coincides with $/$. 
Bonus. The operation $/$ is commutative if and only if it is associative. 
A: On natural numbers, the bitwise XOR operation is commutative, and is its own inverse operation (the neutral element is$~0$).
A: Note that $-$ (minus) is really just a short way of writing $+$ something negative. 
In fact, what you call $\circ^{-1}$ is just the composition $$G \times G \xrightarrow{id \times inv} G \times G \xrightarrow{\cdot (mult)} G$$
So the condition you are asking for $a \circ^{-1} b= b \circ ^{-1} a$  is equivalent to the condition $a\circ b^{-1}=b \circ a^{-1}$. 
But this is equivalent to $a = b \circ a^{-1} \circ b$. If you demand that $G$ is commutative, then this is equivalent to $a^2=b^2$, which for example is true if all elements have order $2$. 
EDIT As Klaus Draeger points out below, the implication that all elements have order two does not need commutativity (see his comment). But then again, if all elements have order two, the group must be commutative...
A: If we define $\circ^{-1}$ for a group, we have a neutral element $1$. Applying commutativity with b = 1 we get
$$1\circ^{-1} a = a \circ^{-1} 1$$
but this simplifies to
$$a^{-1} = a$$
so each element is its own inverse.
A: Part of the problem here is the difference between a function and an operation.
A function f(x) is defined as the mapping from objects x to their counterparts f(x).
The inverse function f-1(x) is the mapping that takes the objects back to where they started.
So f(f-1(x))=x
Under this definition, “adding” is not a function, but “adding two” is a function.
I believe that what you want is what I would describe as a binary operation. It is effectively a function of two values.
So the adding function can be written as add(x, y).
In this description commutativity means that add(x, y) = add(y, x)
I'm not sure that a binary operation can have an inverse in the proper sense, because how can we undo the addition to get the original pair of values? Given that add(x, y) = 6, for example, we can never determine which values x and y were added together.
So let's try to rephrase the question. I started thinking about addition and its “inverse” (or perhaps better “reverse”) subtraction.
Subtraction can also be written as subtract(x, y) and it is not commutative; subtract(x, y) is not equal to subtract(y, x).
The effect of subtraction is really to reverse the addition process.
So subtract(add(x,y), y) = x
The question is therefore if there are a pair of functions f and g that act on two values that have this property:
f(g(x, y), y) = f(g(x, y), x)
But if f is the reverse of g, that would mean that f(g(a, b), b) = a
So we would end up with y = x
Therefore the answer to your question is, “No.”
