A rational function that is its own inverse

The function $f$ is defined by $f:x\mapsto\dfrac{ax}{bx-a}$ for $x\in\mathbb R$, $x\ne\dfrac ab$, where $a$ and $b$ are non-zero constants.

(i) Find $f^{-1}(x)$. Hence or otherwise find $f^2(x)$ and state the range of $f^2$.

(ii) The function $g$ is defined by $g:x\mapsto\dfrac1x$ for all real non-zero $x$.

State whether the composite function $fg$ exists, justifying your answer.

(iii) Solve the equation $f^{-1}(x)=x$.