Is there a function $f$ such that $f^ {-1} (x) = (f(x)) ^ {-1}$?

Question

Let $f(x)=y$, then$f^ {-1} (y) = x$ and $(f(x)) ^ {-1} = \frac{1}{f(x)}$

Answer 1

Here's my (amateur) answer:

if $f^{-1}(x)=\frac{1}{f(x)}$, then $f\left(\frac{1}{f\left(\frac{1}{f(x)}\right)}\right)=\frac{1}{f(x)}$.

But $f\left(\frac{1}{f(x)}\right)=x$. Therefore $f\left(\frac{1}{x}\right)=\frac{1}{f(x)}$.

So $f(x)$ is in the class of functions such that $f\left(x\right)=\frac{1}{f\left(\frac{1}{x}\right)}$.