Can there be an injective function whose derivative is equivalent to its inverse function?

It is possible! Here is an example on the domain $D=[0,\infty)$: $$ f(x) = \bigg(\frac{\sqrt{5}-1}{2}\bigg)^{(\sqrt5-1)/2} x^{(\sqrt5+1)/2}. $$ I found this by supposing that $f(x)$ had the form $ax^b$, setting the derivative equal to the inverse function, and solving for $a$ and $b$.

On $(0, \infty)$, take $f(x) = a x^p$ where $p = (\sqrt{5}+1)/2$ (so that $p(p-1) = 1$) and $a = p^{-1/p}$.

There have already been examples with $f: D \to \mathbb R$, but note that it is not possible with $f:\mathbb R \to \mathbb R$. A simple argument is that for a function $f$ to be injective, necessarily $f'(x) \geq 0$ or $f'(x) \leq 0$ for all $x$. Thus we can see that for there to be equality between $f'(x)$ and $f^{-1}(x)$, then we must have $f^{-1}(x) \geq 0$ or $f^{-1}(x) \leq 0$ for all $x$.

But this can't happen, because any function defined on $f: \mathbb R \to \mathbb R$ must have its inverse go from positive to negative for some $x$. To confirm this, just look at the fact that the inverse of any horizontal line must cross the x-axis by flipping over the line $y=x$, and then add curves to that line to find that nothing has changed, and it still must cross the x-axis.