I want to find all smooth (that is, infinitely differentiable) convex functions $f \colon (0, \infty) \to [0, \infty)$ with $f(1) = 0$ satisfying the functional equation $$ f(x) = x f\left(\frac{1}{x}\right), \qquad x > 0, $$ that is, $f$ should be equal to its own perspective function.
So far, I have only found one example: $f(x) = (x - 1) \ln(x)$ (a non-smooth but convex example is $f(x) = | x - 1 |$). If one drops the smoothness assumption, one can just define a convex function $\phi$ on $(0, 1)$ (e.g. $x \mapsto x^{-\alpha} - 1$ for $\alpha > 0$) and define $$f(x) = \begin{cases} \phi(x), & \text{if } 0 < x < 1, \\ x \phi\left(\frac{1}{x}\right), & \text{if } x > 1 \end{cases}.$$
I noticed that the functional equation is invariant under the scaling $f \mapsto a f$ for $a > 0$ (and the inversion $x \mapsto \frac{1}{x}$), but not under shifts $f \mapsto f + a$ or translations and dilations ($f \mapsto f(\cdot + a)$, $f \mapsto f(a \cdot)$).
Motivation. The entropy function $f$ generates the $f$-divergence between measures $$ D_f(\mu \mid \nu) := \int_{X} f \circ \rho \, \textrm{d}\nu + f_{\infty}' \mu_s(X), $$ where $\mu = \rho \nu + \mu_s$ is the Lebesgue decomposition of $\mu \in M_+(X)$ with respect to $\nu \in M_+(X)$.
The function $f^{\dagger}(x) := x f\left(\frac{1}{x}\right)$ is called Csiszar dual of $f$ and we have $D_f(\mu \mid \nu) = D_{f^{\dagger}}(\nu \mid \mu)$. Since $(f^{\dagger})^{\dagger} = f$, I am interested in all $f$ that are self-dual.
Theorem: A function $f$ on $(0, \infty)$ satisfies $f(x) = xf(1/x)$ if and only if there is some $h$ with $h(x) = h(1/x)$ so that $f(x) = \left(\frac{x}{1+x}\right)h(x)$.
Proof: If $f$ has this form, you can directly compute that $f(x) = xf(1/x)$. For the converse, let $h(x) = \left(\frac{1+x}{x}\right) f(x)$ and show that $h$ satisfies $h(x) = h(1/x)$.
To find such $h$, you may take any even function $g_e(x)$ and set $h(x) = g_e(\log x)$. You can find even functions by choosing any polynomial or power series with even powers, or by taking $g_e(x) = \frac{g(x) + g(-x)}{2}$ to be the even part of any function $g$ on $\mathbb{R}$. We also have $f(1) = 0$ if and only if $g_e(0) = 0$.
We could also substitute $x/(1+x)$ with any $r(x)$ so that $r(x) \neq 0$ for all $x > 0$ and $r(x) = x r(1/x)$; for example, $r(x) = \sqrt{x}$ as mentioned in the OP. Regarding convexity - I'll leave this for you to work out, but presumably you can come up with an appropriate condition on $h$ for $f$ to be convex by computing $f''$ in terms of $h$ and setting $f''(x) > 0$.