Does the functional equation $f(1/r) = rf(r)$ have any nontrivial solutions besides $f(r) = 1/\sqrt{r}$?

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.

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 $\mathbb{R}^+$.

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 $0 < Re(s) <1$