TITLE: For all $x,y\in G$ we have: $f(xf(y))=f(x)y$. Prove that $f$ is an isomorphism?
QUESTION [8 upvotes]: Let $G$ be a group and $f : G \to G$ a function such that for all $x,y\in G$:
$$f(x f(y)) = f(x) y.$$

Prove that $f$ is an isomorphism.

There are two problems here: we don't know that $f$ is a group homomorphism, and we don't know that it is bijective, the two conditions required for $f$ to be an isomorphism.
It is difficult to solve this problem because we don't know the formula for $f$, only an equation it satisfies. Directly proving that the kernel is trivial, or that every element is in the image will be difficult.

REPLY [11 votes]: Note first that $f(e) = e$.
This is because $$f(f(e)) = f(e f(e)) = f(e) e = f(e),$$ and then also $$f(f(e)) = f(f(f(e))) = f(e f(f(e))) = f(e) f(e) = f(e)^{2}.$$
Then show that $f \circ f = 1$, the identity map.
This is because $f(f(y)) = f( e f(y)) = f(e) y = e y = y$.
Thus $f$ is bijective, and a homomorphism, as
$$
f(x y) = f(x f(f(y))) = f(x) f(y).
$$