TITLE: Why $f$ is an automorphism?
QUESTION [10 upvotes]: Let $G$ be  a group and $f: G\rightarrow G$ is a function such that  for any $x,y\in G$ we have 
$$f(xf(y))=f(x)y.$$
Then prove that  $f$ is an automorphism of $G$.

REPLY [12 votes]: By setting $x=1$ we see that $f\circ f$ is the map $y\mapsto f(1)y$, which is bijective.  Therefore $f$ is also bijective.  
From $f(f(1))=f(1)$ and injectiveness we have $f(1)=1$.  It follows that $f\circ f$ is the identity map (in other words, $f=f^{-1}$).  It remains to be shown that $f$ is a homomorphism.
For all $a,b\in G$, $f(ab)=f(af(f(b)))=f(a)f(b)$.