What is an intuitive explanation for $T = \mathrm{const}$ when $\Omega(E) = e^E$? Temperature is related to number of microstates as follows:
$$
\frac{1}{k_{\mathrm{B}}T} = \frac{\mathrm{d}\ln{\Omega(E)}}{\mathrm{d}E} \ .
$$
Hence, if $\Omega(E) = e^E$, then $T = \mathrm{const}$. This contradicts my intuition. I expect temperature of a system to increase as the number of microstates increases. Everything works fine for power functions. For example, if $\Omega(E) = E^x$, then
$$
T = \frac{E}{xk_\mathrm{B}} \ .
$$
I don't understand why exponential function yields such a counterintuitive result, considering it grows faster than power functions.
 A: I'd say the relevant quantity here is $\omega (E):=\ln\,(\Omega (E)) = - \ln \,(1/\Omega(E))$. You can view $\omega$ as a measure of information or rather of missing information, in the sense of how uncertain you are about the exact microstate your system is in (which is the same as the amount of information you gain when you learn about the specific microstate). The (inverse) temperature is a measure of how your knowledge /uncertainty of the specific microstate changes if the energy of the system is changed.
As you've noticed, for $\Omega_1(E)\propto e^{E}$ the number of possible microstates grows faster with respect to energy compared to $\Omega_2(E) \propto E^x$ for some $x\geq 1$ (here and in the following above some threshold energy). Now think about the probability that your system is in a certain microstate, which here is simply $p=1/\Omega(E)$. We then find $p_1 (E)\leq p_2(E)$. So when you learn about the specific microstate, you gain more information, which is why $w_1(E) \geq w_2(E)$ makes sense, too.
You can now ask how your uncertainty about the specific microstate is changed if you change the energy by 'a tiny bit' - that's basically the definition of (the inverse) temperature. For $\omega_1(E)$, you find that this does not change with respect to energy, i.e. it does not matter at which energy scale you change the energy a bit, the uncertainty change is always the same, i.e. $T=\mathrm{const}$. For $\omega_2(E)$, this is not the case. The uncertainty about your microstate changes depending on where on the energy scale you change the energy a bit: The higher the energy, the smaller the change in $\omega_2(E)$ (just compare to a plot of $\ln$). For example, if you double your energy, then $\omega_1(2E) = 2 \omega_1(E)$, i.e. you uncertainty has doubled, but $\omega_2(2E) = \mathrm{const.} + \omega_2(E)$, i.e. the uncertainty is only changed by some fixed amount.
To summarize: $\Omega_1$ grows faster than $\Omega_2$ and hence also $\omega_1$ compared to $\omega_2$, i.e. you know less about the specific microstate at a certain energy. Furthermore, we have $T_2(E) \geq T_1(E)$. Again: That $T_1(E)$ is constant just means that irrespective of how large or small the energy of your system is, your change in uncertainty if you change the energy is always the same. In contrast, $T_2(E) \propto E$ means that the larger the energy, the larger the temperature and hence the smaller the change in your uncertainty about the microstate if you change the energy (infinitesimally).
A: The basic intuition here is that temperature is not about number of microstates as such. Rather, it is about how the number of microstates varies with the energy---the standard definition of temperature in terms of entropy and energy.
A hot system does not need to have a lot of microstates. The white-hot pieces of metal leaping off a firework 'sparkler' have less entropy than a pool of water freezing on a cold night.
The dependence $S \propto E$ is unusual because it has zero second derivative. Such an entropy function is neither convex nor concave, so it is on the boundary of the stability criterion, which means it is unstable. In other words a system with this entropy function cannot be in a stable equilibrium. In practice the dependence $S \propto E$ will not extend to arbitrarily high energies; it will curve over and then the whole function is somewhat like the one describing a phase transition.
