We can expect the findings you observed because the heat flux at any point depends on conduction and radiation (and thus depends on the position, emissivity, thermal conductivity, and surrounding temperature), whereas the heat flux per temperature gradient at any interior point depends only on the thermal conductivity, per Fourier's law.
To back out the thermal conductivity from the heat flux alone, one needs to solve the heat equation within the material for the entire problem.
Let's consider an example: Idealize the material as long and narrow, to permit a 1-D approximation in the $x$-direction. A steady-state energy balance of any slice of the material yields
$$kA\frac{d^2 T(x)}{dx^2}=\sigma\varepsilon P [T(x)^4-T_\infty^4],$$
with thermal conductivity $k$, cross-sectional area $A$, temperature $T$, Stefan–Boltzmann constant $\sigma$, emissivity $\varepsilon$, perimeter length $P$ (=$2A/r$ for a circular cross section with radius $r$) and surrounding enclosure temperature $T_\infty$. That is, the conduction input balances the radiation output.
This nonlinear PDE would need to be solved numerically. If $T(x)$ isn't much different from $T_\infty$, though, we can linearize the radiation term as approximation $4T_\infty^3 T^\prime(x)$, where $T^\prime(x)=T(x)-T_\infty$. With this approximation, we have a linear PDE:
$$\frac{d^2 T^\prime(x)}{dx^2}-C_1T^\prime(x)=0,$$
where $C_1=\frac{4\sigma\varepsilon PT_\infty^3}{kA},$ with solution
$$T^\prime(x)=C_2\sinh\left(\sqrt{C_1}x\right)+C_3\cosh\left(\sqrt{C_1}x\right)$$
subject to boundary conditions $\left.\frac{dT^\prime(x)}{dx}\right|_0=\frac{q_\mathrm{left}}{k}$ and $T^\prime(x)|_L=T_\mathrm{right}-T_\infty$ (with length $L$), so
$$T(x)=T_\infty+\frac{q}{k\sqrt{C_1}}\sinh(\sqrt{C_1}x)+\frac{1}{\cosh(\sqrt{C_1}L)}\left(T_\mathrm{right}-T_\infty-\frac{q}{k\sqrt{C_1}}\sinh(\sqrt{C_1}L)\right)\cosh(\sqrt{C_1}x),$$
from which one could back out the thermal conductivity if they knew the temperature and the other material properties.
Clearly, multiple assumptions and idealizations have been needed here to obtain a closed-form expression, but I hope this clarifies why the thermal conductivity isn't straightforwardly linked with the temperature $T$ or temperature gradient $\nabla T$ alone but is straightforwardly linked with the conductive heat flux $-k\nabla T$ per unit gradient $\nabla T$, again per Fourier's law.
