The Shannon entropy of a discrete probability distribution $\newcommand{\bs}[1]{\boldsymbol{#1}}\bs p\equiv (p_i)_{i=1}^n$ is defined as $H(\bs p)\equiv -\sum_{i=1}^n p_i \log p_i$.
Consider the corresponding level sets, that is, the sets of the form $$L^{(n)}_\alpha\equiv \left\{(p_1,...,p_n) : \sum_i p_i=1 \text{ and } H(p_1,...,p_n)=\alpha\}\subset\mathbb R^n\right\},\quad\alpha\in[0,\log n].$$
Is there a geometrical characterisation for these sets?
Clearly, $L^{(n)}_{\log n}=\{(1,...,1)/n\}$ and $L^{(n)}_{0}=\{\bs e_1,...,\bs e_n\}$ where $(\bs e_i)_j=\delta_{ij}$. What about the nontrivial cases with $0<\alpha<\log n$?
For example, in the $n=3$ case the corresponding level sets/contour lines look like in the following:
To get a better look at the contour lines we can parametrise the simplex as $$S(s,t)=(1,0,0)+\frac{t}{\sqrt2}(-1,1,0)+\frac{s}{\sqrt{3/2}}(-1/2,-1/2,1),$$ and then plotting $H(S(s,t))$ against $s,t\in\mathbb R^2$ we get
We can push this further to visualise single level sets for $n=4$, by using the parametrisation $$S(s,t,u) = (1,0,0,0) + \frac{t}{\sqrt2}(-1,1,0,0) + \frac{s}{\sqrt{3/2}}(-1/2,-1/2,1,0) + \frac{u}{\sqrt{4/3}}(-1/3,-1/3,-1/3,1),$$ and them plotting the $(s,t,u)$ such that $H(S(s,t,u))=\alpha$. For example, with $\alpha=\log(3.2)$ we get
where the tetrahedron shows how the normalisation constraint on the probabilities is translated into this $(s,t,u)$ space.
The fact that $H$ doesn't care about the ordering of the elements in $\bs p$ implies a series of reflection symmetries on the level sets. What else can be said about them?
The fact that not all such level sets are closed might make the problem less well-defined, in which case we might restrict our attention to the cases with $\log(n-1)\le \alpha \le \log n$ for which (I think) the level sets should be closed. Alternatively, one might extend the definition of $H$ to let it act on vectors that are not necessarily probability distributions.
The Mathematica code to generate the figure can be found here.
Only an answer to Arnaud Mégret's comment. Close to the max-entropy point, the region indeed tends to a sphere.
Let $p_i=\frac{1}{n}+\epsilon_i$ Then we do a Taylor expansion of the entropy for small $\epsilon_i$ to get (using natural logarithms, i.e. , nats):
$$H \approx \log(n) + (\log(n)-1) \sum_i \epsilon_i - \frac{n}{2} \sum_i\epsilon_i^2 = \log(n) - \frac{n}{2} \sum_i\epsilon_i^2 $$
The linear term vanishes because of the restriction $\sum_i \epsilon_i =0$. Hence, in the vecinity of the max-entropy point (uniform distribution), the constant-entropy regions have the approximate form
$$ \sum_i \epsilon_i^2 = \sum_i (p_i - 1/n)^2=r^2=\frac{2}{n}(\log(n)-H)$$
That is, a sphere - more precisely, a $n$-sphere intersected with a plane, which is a $n-1$-sphere.