Does the functional equation $f(1/r) = rf(r)$ have any nontrivial solutions besides $f(r) = 1/\sqrt{r}$? Repeating for the sake of TeX rendering:
Does the functional equation $f(1/r) = rf(r)$ have any nontrivial solutions besides $f(r) = 1/\sqrt{r}$?
 A: Yes. If you want another function defined on $\mathbb{R}^+$:
Suppose you have a function $g$ which is invariant under inversion $g(1/z)=g(z)$,
then $f(z)\cdot g(z)$ is a new function satisfying your functional equation.
(f is your function $z\mapsto 1/\sqrt{z}$)
For $g$ you can for example take $z\mapsto ln(z)^2$.
EDIT: I just wanted to add a "full solution" to this problem.
Suppose you have a function f on $\mathbb{R}^+$ which satisfies your functional equation:
Define $g(r) = \frac{1}{2}\sqrt{r}f(r)$, which is invariant under inversion and we have 
$f(r)=\frac{2}{\sqrt{r}}\cdot g(r)=\frac{1}{\sqrt{r}}\cdot(g(r)+g(\frac{1}{r}))$.
Conversely for any function g on $\mathbb{R}^+$ we have that $f=\frac{1}{\sqrt{r}}\cdot(g(r)+g(\frac{1}{r}))$ satisfies your functional equation.
So the set of solutions to your equation is $\{\frac{1}{\sqrt{r}}\cdot(g(r)+g(\frac{1}{r}))\}$, where g runs through all the functions on $\mathbb{R}^+$.
A: Jacobi's theta function, which is intimately related to the Riemann zeta function, is a very famous solution of your functional equation.
The functional equation is related to symmetry properties of the Mellin transform:
For $x>0,$ given the Mellin transform for $0<real(s)<1$
$$\hat{f}(s)=\int_{0}^{\infty }x^{s-1}f(x)\,dx,$$ then
$$ f(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \hat{f}(s) x^{-s}ds$$
for $0 < \sigma <1$, and 
$$f(x)=\frac{1}{x}f(\frac{1}{x})\text{ iff } \hat{f}(s)=\hat{f}(1-s).$$
(Two changes of variables show this:  $x$ to $1/x$ and $s$ to $1-s$.)
Riemann's $\xi(s)$ function satisfies, for $0<real(s)<1$,
$$\xi(s)=\xi(1-s) = \pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s)=\int_{0}^{\infty }[\vartheta (0;ix^2)-(1+\frac{1}{x})]x^{s-1}\,dx.$$
where  $\vartheta (z;\tau)$ is Jacobi's theta function. 
By the Mellin transform theorem then
$$\psi(x)=\vartheta (0;ix^2)-(1+\frac{1}{x})=\frac{1}{x}\psi(\frac{1}{x})$$ 
and since $$g(x)=1+\frac{1}{x}=\frac{1}{x}g(\frac{1}{x}),$$ then 
$$\vartheta (0;ix^2)=\frac{1}{x}\vartheta (0;\frac{i}{x^{2}}).$$
Note: As a series expression
$$\psi(x)=\frac{-1}{x}+2\sum_{n=1}^{\infty}exp(-\pi n^{2}x^2),$$
and for $0<real(s)<1$
$$\xi(s)=\lim_{L\to +\infty, a\to 0^+}\frac{L^{s-1}}{s-1}+\frac{-a^{s}}{s}+\int_{a}^{L }[\vartheta (0;ix^2)-(1+\frac{1}{x})]x^{s-1}\,dx.$$
Added Oct. 4, 2019:
From the arguments above, any function $\hat{f}(s)$ we construct that is symmetric through the line $ Re(s) = 1/2$ has the symmetry $ \hat{f}(1-s)=\hat{f}(s)$ and its inverse Mellin transform if it exists for $0 < \sigma< 1$ will give us a function such that $ f(x) = \frac{1}{x}f(\frac{1}{x})$.
For example:
1) $\hat{f}(s) = \frac{1}{s} + \frac{1}{1-s}$ and $f(x)= H(1-x) 1+H(x-1)\frac{1}{x}$ for $\sigma = 1/2$ for our line of integration, where $H(x)$ is the Heaviside step function. Note $1+ \frac{1}{x}$ is also a solution but has no Mellin transform.
2) $\hat{f}(s) = (s-1)! + (-s)!$ and $f(x)= e^{-x} + \frac{1}{x}e^{-\frac{1}{x}}$.
3) and one of the most important "functions", the Dirac delta function, has this property
$ \delta(x-1)= \frac{1}{x}\delta[\tfrac{1}{x}-1]$
with
$\widehat{\delta}(s) = 1 = \widehat{\delta}(1-s)$.
Clearly, the pattern 
$$f(x)= w(x) + \frac{1}{x}w(\frac{1}{x})$$
satisfies the basic functional relation for any function $w(x)$ and also satisfies the Mellin space relation if $w(x)$ is transformable in the strip $0 < Re(s) <1$
A: There are a huge number of solutions. Let $g$ be any function from $(-1,1]$ to $\mathbb{R}$. Define a function $f: \mathbb{R} \to \mathbb{R}$ by $f(x)=g(x)$ if $x \in (-1,1]$, $f(-1)=0$, and $f(x)=(1/x)g(1/x)$ otherwise. Then $f$ satisfies your equation.
