Romanian Mathematical Olympiad

Assume, by way of contradiction, that there exist $a, b \in G$ such that $f(a) = a \neq a^{-1}$ and $f(b) = b^{-1} \neq b$. Then $f(ab) = f(a)f(b) = ab^{-1} \neq ab$, therefore $f(ab) = (ab)^{-1} = b^{-1}a^{-1}$. It follows that $ab^{-1} = b^{-1}a^{-1}$, which implies $b^{-1} = ab^{-1}a$.

Next, $f(ab^2) = f(a)f^2(b) = ab^{-2}$. If $f(ab^2) = ab^2$, then $ab^{-2} = ab^2$, therefore either $b^2 = e$, which contradicts $b \neq b^{-1}$, or $\text{ord}(b) = 4$, which is impossible. Thus, $f(ab^2) = (ab^2)^{-1} = b^{-2}a^{-1}$, hence $ab^{-2} = b^{-2}a^{-1}$. It follows that $ab^{-2}a = b^{-2} = (b^{-1})^2 = ab^{-1}aab^{-1}a$, so $a^2 = e$, again a contradiction.