linear algebra - When are almost block companion matrices which yield a given characteristic polynomial connected?

This is motivated by this question Are matrices which yield a given characteristic polynomial and have specified structure connected?

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Let $\mathcal E \in M_n(\mathbb R)$ be a subset with following form: we first construct a block diagonal matrix in $M_n(\mathbb R)$ such that
\begin{align*}
C = \begin{pmatrix}
C_{k_1} & 0 & 0 & \cdots & 0 \\
0 & C_{k_2} & 0 & \cdots & 0 \\
0 & 0 & \ddots & \vdots & 0 \\
0 & 0 & 0 & \cdots & C_{k_r}
\end{pmatrix},
\end{align*} with $k_1 + k_2 + \dots + k_r = n$
such that each block $C_{k_j}$ is in the companion form

\begin{align*}
\begin{pmatrix}
0 & 0 & \cdots & 0 & -c_0 \\
1 & 0 & \cdots & 0 & -c_1 \\
0 & 1 & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1& -c_{k_j-1}
\end{pmatrix}.
\end{align*}
Now for each block we extend the last column to fill up the whole matrix. For example, suppose we have two blocks $C_1$ and $C_2$ with $C_1 \in \mathbb R^{2 \times 2}$ and $C_2 \in \mathbb R^{3 \times 3}$, elments in $\mathcal E$ would look like
\begin{align*}

\begin{pmatrix}
0 & -a_1 & 0 & 0 & -b_1 \\
1 & -a_2 & 0 & 0 & -b_2 \\
0 & -a_3 & 0 & 0 & -b_3 \\
0 & -a_4 & 1 & 0 & -b_4 \\
0 & -a_5 & 0 &1 & -b_5
\end{pmatrix}.
\end{align*}

It is also clear for any monic $n^{th}$ degree real polynomial, we can at least find one realization in $\mathcal E$ since we can choose a matrix in block diagonal form. Let $f: \mathcal E \to \mathbb R^n$ be the map sending the coefficients of characteristic polynomial to $\mathbb R^n$.

Let $q(t) = t^n + a_{n-1} t^{n-1} + \dots + a_0$ be a fixed polynomial. I am wondering whether there are sufficient conditions on $a = (a_{n-1}, \dots, a_0)$ such that $f^{-1}(a)$ is a connected set?

This question Are matrices which yield a given characteristic polynomial and have specified structure connected? asked a specific case, i.e, $n=4, k_1 = k_2 = 2$. There is a very nice answer proving: as long as the polynomial has a real root, then it is connected. The technique by the answer does not seem to generalize. But I am very interested to know whether the same condition holds here: if $q(t)$ has a real root, then $f^{-1}(a)$ is connected where $a = (a_{n-1}, \dots, a_0)$ is the coefficient vector of $q(t)$?

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EDIT: This question might be too tricky to answer (This is the third time I put a bounty). But I would be happy to reward the bounty if someone gives an answer on a very special polynomial with coefficients vector of $a$ such that $f^{-1}(a)$ is connected. For example, is $f^{-1}((0, \dots, 0))$ connected, i.e., the polynomial with all roots to be $0$ or some other special polynomials?