I now have a condition which I believe is necessary and sufficient in the special case of a finite sample space. My attempted proof is below.
Claim
Let $\Omega = [n]$, equipped with probability $Q=(q_1,\ldots,q_n)$ in the probability simplex in $R^n$. For a given risk metric $\rho:R^n \rightarrow R$, a latent disutility function $f:R \rightarrow R$ exists satisfying $\rho(Z)=E(f \circ Z)$ if and only if for every $\{x_1, \ldots, x_n\} \subseteq R$, $${\scr Q{\bf f}=R},$$ where $\scr Q$ is an $\underbrace{n \times \cdots \times n}_\text{$n+1$ times}$ tensor, ${\bf f} \in R^n$, and $\scr R$ is an $\underbrace{n \times \cdots \times n}_\text{$n$ times}$ tensor, with the components
$${\scr q}_{\omega_1 \ldots \omega_n k}=\sum_j q_j I(\omega_j=k)$$
$${\bf f}_\omega = \rho(x_\omega, \ldots, x_\omega)$$
$${\scr r}_{\omega_1 \ldots \omega_n}=\rho(x_{\omega_1}, \ldots, x_{\omega_n}).$$.
Proof
$(\implies)$ Assume such an $f$ exists. A cost variable $Z=(z_1,\ldots,z_n)$ is an element of $R^n$, and $E(Z) =\sum_j q_j z_j$. If $Z$ is constant, then $z_j = c$ for each $j \in [n]$, and thus $$E(f \circ Z) =\sum_j q_jf(z_j)= \sum_j q_jf(c)=f(c).$$
Hence $f(c) = \rho(c,\ldots,c)$ for each $c \in R$.
For any $\{x_1, \ldots, x_n\} \subseteq R$, there are $n^n$ random variables supported on the $x_i$, each of which has the form $Z = (x_{\omega_1}, \ldots, x_{\omega_n})$. Each such $Z$ places a linear constraint on $\rho$: $$\rho(x_{\omega_1}, \ldots, x_{\omega_n})=\rho(Z)=E(f \circ Z)=\sum_j q_jf(x_{\omega_j})=\sum_j q_j\rho(x_{\omega_j}, \ldots, x_{\omega_j}).$$ The equation ${\scr Q{\bf f}=R}$ expresses all of these constraints in tensor form, so it must hold.
$(\impliedby)$ From above, $f:R \rightarrow R$ defined by $f(c) = \rho(c,\ldots,c)$ has the desired property. This completes the proof.
Discussion
This result allows one to easily reject risk functions which do not have associated latent disutilities, by evaluating $\rho$ for particular $Z$ and enforcing linear constraints. It also provides an explicit construction of the unique $f$ from $\rho$ in cases where such an $f$ exists. This makes it easy to check monotonicity/continuity/convexity.
Unfortunately, even though the condition I present is sufficient, it would be quite difficult to use this result to confirm risk functions which do have associated latent disutilities. I'd love to see if someone who knows more about crazy linear algebra could find a more easily verifiable sufficient condition. And of course, a generalization to infinite sample spaces would be incredible, although at this point I'm afraid I wouldn't understand it without a PhD in real analysis.
Best Answer
Let $\omega \in \Omega$, define a risk measure $\rho$ like:
$$ \rho(Z) = Z(\omega) $$
$Z \in \mathcal{Z}=\mathcal{L}(\Omega, \mathcal{F}, P)$ is a random variable. The risk measure $\rho$ is well-defined in this way.
Let $Z_1 \sim \mathcal{N}(0, 1), Z_2=-Z_1$, we know that $Z_1$ and $Z_2$ has the same distribution but generally $\rho(Z_1)\neq \rho(Z_2)$.