Function satisfying $f^{-1} =f'$
Seems like we have here yet another vindication of Léo Sauvé's famed dictum...
I am going to try to sketch below the awe-inspiring solution to this problem with which A. C. Hindmarsh came up and which was showcased on page 696 of the sixth issue of volume 76 (1969) of the American Mathematical Monthly. As far as I know, the problem was originally proposed by H. L. Nelson and appeared on page 779 of volume 75 (1968) of the Monthly: oddly enough, it would make its way to the problem & solutions column of the Monthly once again in 1976 (vol. 83, no. 4, p. 293); however, on that occasion it would be attributed to a Nathaniel Grossman from UCLA.
Solution. We wish to determine all functions $f \colon [0,\infty) \to [0,\infty)$ such that $f(0)=0$ and $f^{\prime}(x) = f^{-1}(x)$ for every $x \in I:=(0,\infty)$. We shall prove that, apart from the function mentioned by Tsuyoshi Ito in his reply, there is no other function $f$ which satisfies all the constraints under consideration.
First off, we notice that if $f$ is a function that does the job, then $f$ must be $\mathcal{C}^{1}$ and strictly increasing in $(0,\infty)$. Then, differentiating the identity $$f(f^{\prime}(x))=x$$ repeatedly, we obtain that $f$ is a function of class $\mathcal{C}^{\infty}$. What is more, we obtain that $f^{\prime \prime}>0$, $f^{\prime \prime \prime} < 0, \ldots, (-1)^{k}f^{(k)}>0$; it follows from Bernstein's theorem on regularly monotonic functions that $f$ is a real-analytic function on $(0,\infty)$.
Now, from the identity
$$\frac{d}{dx} f(f(x)) = f^{\prime}(f(x))f^{\prime}(x) = xf^{\prime}(x)$$
we get that
$$f(f(x)) = \int_{0}^{x} y \, f^{\prime}(y) \, dy$$ for every $x \in I$. This allows us to ascertain that $f$ has a fixed point $a \in I$: if this were not the case, the function $F \colon I \to \mathbb{R}$, defined for every $x \in I$ as $F(x)= f(x)-x$, would be of fixed sign. We claim that such a thing is not possible: indeed, if $f(x)>x$ for every $x \in I$, then $y = f^{\prime}(f(y)) > f^{\prime}(y)$ for every $y \in (0,x)$ and whence
$$ x < f(f(x)) = \int_{0}^{x} y \, f^{\prime}(y) \, dy < \int_{0}^{x} y^{2} \, dy = \frac{x^{3}}{3},$$
which doesn't necessarily hold when $x$ is sufficiently small; since the assumption that the inequality $f(x)<x$ holds for every $x \in I$ allows us to derive a similar contradiction, we conclude that any solution $f$ to the functional-differential in question does have a fixed point $a\in I$. Further, the strict convexity of $F$ implies that $F$ has at most two zeros, counting the one it has at $x=0$. Thus, $f$ has exactly one fixed point $a\in I$, with $f(x)<x$ in $(0,a)$, $f(x)>x$ in $(a,\infty)$, $f^{\prime}(x)>x$ in $(0,a)$, and $f^{\prime}(x)<x$ in $(a,\infty)$.
The uniqueness of the solution to the problem is established by means of the fixed point whose existence has just been proven. Let us suppose that $f_{1}$ and $f_{2}$ are two functions satisfying all the constraints under consideration and let $g:=f_{1}-f_{2}$. Moreover, let us denote with $a_{i}$ the unique fixed point of $f_{i}$ in the interval $I$. Without loss of generality, we can suppose that $a_{1} \geq a_{2}$.
The possibility that $a_{1}>a_{2}$ leads us to a contradiction. (I am not adding all the details behind the corresponding analysis for the moment. I consider this is the least transparent part of Mr. Hindmarsh's argument. Can any of you, fellow MO-ers, provide a more limpid justification of this part of the solution?)
If $a_{1}=a_{2}=a$, then it is not difficult to convince oneself that $0=g(a)=g^{\prime}(a) = g^{\prime \prime}(a) =\ldots$; being $g$ a real-analytic function in $I$, the latter equalities implies that $g$ vanishes identically and we are done.
QED.
Wow. I remember that I thought exactly the same problem out of curiosity as a high school student but did not reach an answer. In fact, I was thinking about posting this problem on MathOverflow!
At least it is easy to construct one solution: f(x)=xφ/φφ−1, where φ=(1+√5)/2 is the golden ratio.
Edit: Corrected the calculation. Thanks to Aaron Meyerowitz for spotting the error!
Let $a=1+p>1$ be given. We shall construct a function $f$ of the required kind with $f(a)=a$ by means of an auxiliary function $h$, defined in the neighborhood of $t=0$ and coupled to $f$ via $x=h(t)$, $f(x)=h(a t)$, $f^{-1}(x)=h(t/a)$. The condition $f'=f^{-1}$ implies that $h$ satisfies the functional equation $$(*)\quad h(t/a) h'(t)=a h'(at).$$ Writing $h(t)=a+\sum_{k \ge 1} c_k t^k$ we obtain from $(*)$ a recursion formula for the $c_k$, and one can show that $0< c_r<1/p^{r-1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $|t|< p$, satisfies $(*)$ and possesses an inverse $h^{-1}$ in the neighborhood of $t=0$. It follows that the function $f(x):=h(ah^{-1}(x))$ has the required properties.