The first thing to notice is that each of these measures is in opposite directions, and they are also on different scales. In order to compare them in the same direction and scale, I am going to compare scaled versions of the negated HHI and entropy. Specifically, I will begin by comparing the following functions:
$$\begin{aligned}
R(\mathbf{p}) &\equiv \frac{n-1}{n} \bigg( 1 - \sum_{i=1}^n p_i^2 \bigg), \\[6pt]
S(\mathbf{p}) &\equiv - \frac{1}{\log n} \sum_{i=1}^n p_i \log p_i. \\[6pt]
\end{aligned}$$
The HHI and the entropy are affine transformations of these two functions, so if we compare these two scaled functions, we will get simple corresponding results for the measures of interest. To see why I have chosen to examine these two functions, consider the special input vectors $\mathbf{u} \equiv (\tfrac{1}{n},...,\tfrac{1}{n})$ (all probabilities equal) and $\mathbf{m} \equiv (1,0,...,0)$ (one probability dominating). At these extremes we have the following results:
$$\begin{matrix}
R(\mathbf{m}) = 0 & & & & R(\mathbf{u}) = 1, \\[6pt]
S(\mathbf{m}) = 0 & & & & S(\mathbf{u}) = 1. \\[6pt]
\end{matrix}$$
You can see from the above that the scaled functions I am using range between zero and one; they attain the zero value when one probability dominates the others and they attain unity when all the probabilities are equal. This means that both functions $R$ and $S$ are scaled measures of equality.
Rates-of-change of scaled equality measures: From the above forms of the functions, hopefully you can get a sense of the difference in the scaled measures. Below we will show the rates-of-change of the measures for a change in the probability vector. We will show that increasing a given probability will increase or decrease $R$ depending on whether that probability is below or above the arithmetic mean of the other probabilities. Contrarily, increasing a given probability will increase or decrease $S$ depending on whether that probability is below or above the geometric mean of the other probabilities.
We will examine rates-of-change as we alter one of the probabilities, with corresponding changes in other probabilities. To retain the norming requirement for the probability vector, we will consider that increasing the probability $p_k$ by some small amount $d p$ is accompanied by a corresponding change in all the other probabilties of $- \tfrac{1}{n-1} d p$. Thus, we have:
$$\frac{d p_i}{d p_k} = - \frac{1}{n-1}
\quad \quad \quad \text{for } i \neq k.$$
Using the chain rule for total derivatives, for any $\mathbb{p}$ in the interior of its allowable range we therefore have:
$$\begin{aligned}
\frac{d R}{d p_k} (\mathbf{p})
&= \sum_{i=1}^n \frac{d p_i}{d p_k} \cdot \frac{\partial R}{\partial p_i} (\mathbf{p}) \\[6pt]
&= \frac{\partial R}{\partial p_k} (\mathbf{p}) + \sum_{i \neq k} \frac{d p_i}{d p_k} \cdot \frac{\partial R}{\partial p_i} (\mathbf{p}) \\[6pt]
&= - \frac{n-1}{n} \cdot 2 p_k + \sum_{i \neq k} \frac{1}{n-1} \cdot \frac{n-1}{n} \cdot 2 p_i \\[6pt]
&= - 2 \cdot \frac{n-1}{n} \Bigg[ p_k - \frac{1}{n-1} \sum_{i \neq k} p_i \Bigg], \\[6pt]
\end{aligned}$$
and:
$$\begin{aligned}
\frac{d S}{d p_k} (\mathbf{p})
&= \sum_{i=1}^n \frac{d p_i}{d p_k} \cdot \frac{\partial S}{\partial p_i} (\mathbf{p}) \\[6pt]
&= \frac{\partial S}{\partial p_k} (\mathbf{p}) + \sum_{i \neq k} \frac{d p_i}{d p_k} \cdot \frac{\partial S}{\partial p_i} (\mathbf{p}) \\[6pt]
&= - \frac{1}{\log n} \Bigg[ (1 + \log p_k) - \frac{1}{n-1} \sum_{i \neq k} (1 + \log p_i) \Bigg] \\[6pt]
&= - \frac{1}{\log n} \Bigg[ \log p_k - \frac{1}{n-1} \sum_{i \neq k} \log p_i \Bigg]. \\[6pt]
\end{aligned}$$
We can see that the two measures have different "cross-over points" for when an increase to $p_k$ increases or decreases the measure. For the measure $R$ the cross-over point is where $p_k$ is equal to the arithmetic mean of the other probabilities; below this point, increasing $p_k$ increases the measured equality between the elements and so it increases $R$. For the measure $S$ the cross-over point is where $p_k$ is equal to the geometric mean of the other probabilities; below this point, increasing $p_k$ increases the measured equality between the elements and so it increases $R$.
Relative rates-of-change and limiting cases: Aside from having different "cross-over" points, the two measures also change at different rates relative to one another when we change $p_k$. For a small increase in the probability $p_k$ we have:
$$\frac{dR}{dS} (\mathbf{p}) = \frac{d R}{d p_k} (\mathbf{p}) \Bigg/ \frac{d S}{d p_k} (\mathbf{p}) = \frac{2 (n-1) \log n}{n} \cdot \frac{p_k - \frac{1}{n-1} \sum_{i \neq k} p_i}{\log p_k - \frac{1}{n-1} \sum_{i \neq k} \log p_i}.$$
It is useful to examing this relative rate-of-change in the extreme cases. In particular, we have:
$$\lim_{p_k \uparrow 1} \frac{dR}{dS} (\mathbf{p}) = 0
\quad \quad \quad
\lim_{p_k \downarrow 0} \frac{dR}{dS} (\mathbf{p}) = 2 \cdot \frac{n-1}{n} \cdot \frac{\log n}{\sum_{i \neq k} \log p_i}.$$
This shows that when $p_k$ is a dominating probability, which is near one, increasing it further will decrease $S$ much more rapidly than it decreases $R$. Contrarily, when $p_k$ is a dominated probability, which is near zero, increasing it increases $S$ much more rapidly than it increases $R$, and this is especially pronounced when $n$ is large.
