Okay, I believe, I have constructed the bridge between the adjoint of a vector equation and the adjoint of a scalar equation. I will explain it step by step here for those who might be interested.
Higher-Order Two-Term Scalar Equations
Now, consider the following higher-order two-term scalar differential equation
\begin{equation}
x^{(n)}(t)+p(t)x(t)=0,\label{hottceeq1}\tag{1}
\end{equation}
where $p$ is a complex-valued continuous function.
Then, the matrix representation for \eqref{hottceeq1} is
\begin{equation}
\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right)^{\prime}
=
\left(
\begin{array}{cccc}
&1&&\\
&&\ddots&\\
&&&1\\
-p(t)&&&
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right).\label{hottceeq2}\tag{2}
\end{equation}
Thus, we obtain
\begin{equation}
(-1)
\left(
\begin{array}{cccc}
&1&&\\
&&\ddots&\\
&&&1\\
-p(t)&&&
\end{array}
\right)^{\ast}
=
\left(
\begin{array}{cccc}
&&&\overline{p}(t)\\
(-1)&&&\\
&\ddots&&\\
&&(-1)&
\end{array}
\right)\notag
\end{equation}
and
\begin{equation}
\left(
\begin{array}{c}
-y^{(n-1)}\\
y^{(n-2)}\\
\vdots\\
(-1)^{n}y
\end{array}
\right)^{\prime}
=
\left(
\begin{array}{cccc}
&&&\overline{p}(t)\\
(-1)&&&\\
&\ddots&&\\
&&(-1)&
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
-y^{(n-1)}\\
y^{(n-2)}\\
\vdots\\
(-1)^{n}y
\end{array}
\right),\label{hottceeq3}\tag{3}
\end{equation}
where we have constructed the unknown matrix from bottom to the top.
This system gives us the adjoint equation
\begin{equation}
(-1)^{n}y^{(n)}(t)+\overline{p}(t)y(t)=0.\notag
\end{equation}
Note that, if $n=\text{even}$, then both \eqref{hottceeq2} and \eqref{hottceeq3} represent \eqref{hottceeq1},
i.e., \eqref{hottceeq1} is self-adjoint.
Further, using the definition of the inner product in the first post, we get
\begin{equation}
\left\langle\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right),
\left(
\begin{array}{c}
-y^{(n-1)}\\
y^{(n-2)}\\
\vdots\\
(-1)^{n}y
\end{array}
\right)\right\rangle
=\sum_{j=0}^{n-1}(-1)^{n-j}x^{(n-1-j)}\overline{y}^{(j)}=\text{constant}.\notag
\end{equation}
Higher-Order Autonomous Scalar Equations
Next, consider the
\begin{equation}
x^{(n)}(t)+p_{1}x^{(n-1)}(t)+\cdots+p_{n}x(t)=0,\label{hoaseeq1}\tag{4}
\end{equation}
where $p_{1},p_{2},\cdots,p_{n}$ are complex numbers.
Then, the matrix representation for \eqref{hoaseeq1} is
\begin{equation}
\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right)^{\prime}
=
\left(
\begin{array}{cccc}
&1&&\\
&&\ddots&\\
&&&1\\
-p_{n}&-p_{n-1}&\cdots&-p_{1}
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right).\notag
\end{equation}
On the other hand, by using the adjoint coefficient matrix, we form the differential system
\begin{equation}
\begin{aligned}[]
&\left(
\begin{array}{c}
(-1)^{n-1}y^{(n-1)}+(-1)^{n-2}\overline{p}_{1}y^{(n-2)}+\cdots+\overline{p}_{n-1}y\\
(-1)^{n-2}y^{(n-2)}+(-1)^{n-3}\overline{p}_{2}y^{(n-3)}+\cdots+\overline{p}_{n-2}y\\
\vdots\\
y
\end{array}
\right)^{\prime}\\
=&
\left(
\begin{array}{cccc}
&&&\overline{p}_{n}\\
(-1)&&&\overline{p}_{n-1}\\
&\ddots&&\vdots\\
&&(-1)&\overline{p}_{1}
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
(-1)^{n-1}y^{(n-1)}+(-1)^{n-2}\overline{p}_{1}y^{(n-2)}+\cdots+\overline{p}_{n-1}y\\
(-1)^{n-2}y^{(n-2)}+(-1)^{n-3}\overline{p}_{2}y^{(n-3)}+\cdots+\overline{p}_{n-2}y\\
\vdots\\
y
\end{array}
\right),
\end{aligned}
\notag
\end{equation}
which gives the scalar equation
\begin{equation}
(-1)^{n}y^{(n)}(t)+(-1)^{n-1}\overline{p}_{1}y^{(n-1)}(t)+\cdots+\overline{p}_{n}y(t)=0.\notag
\end{equation}
Further, putting $p_{0}:=1$ for simplicity, we have
\begin{equation}
\left\langle\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right),
\left(
\begin{array}{c}
\begin{array}{c}
\sum_{j=0}^{n-1}(-1)^{j}\overline{p}_{n-1-j}y^{(j)}\\
\sum_{j=0}^{n-2}(-1)^{j}\overline{p}_{n-2-j}y^{(j)}\\
\vdots\\
y
\end{array}
\end{array}
\right)\right\rangle
=\sum_{k=0}^{n-1}\sum_{j=0}^{k}(-1)^{j}p_{k-j}x^{(n-1-k)}\overline{y}^{(j)}=\text{constant}.\notag
\end{equation}
Higher-Order Scalar Equations with Variable Coefficients
Finally, we consider
\begin{equation}
x^{(n)}(t)+p_{1}(t)x^{(n-1)}(t)+\cdots+p_{n}(t)x(t)=0,\notag
\end{equation}
where $p_{i}(t)$ ($i=1,2,\cdots,n$) is complex-valued and is $i$ times continuously differentiable function.
Thus, the matrix representation is
\begin{equation}
\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right)^{\prime}
=
\left(
\begin{array}{cccc}
&1&&\\
&&\ddots&\\
&&&1\\
-p_{n}(t)&-p_{n-1}(t)&\cdots&-p_{1}(t)
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right).\notag
\end{equation}
Then, the associated adjoint matrix is
\begin{equation}
-
\left(
\begin{array}{cccc}
&1&&\\
&&\ddots&\\
&&&1\\
-p_{n}(t)&-p_{n-1}(t)&\cdots&-p_{1}(t)
\end{array}
\right)^{\ast}
=
\left(
\begin{array}{cccc}
&&&\overline{p}_{n}(t)\\
(-1)&&&\overline{p}_{n-1}(t)\\
&\ddots&&\vdots\\
&&(-1)&\overline{p}_{1}(t)
\end{array}
\right),\notag
\end{equation}
which yields the system
\begin{equation}
\begin{aligned}[]
&\left(
\begin{array}{c}
\sum_{j=0}^{n-1}(-1)^{j}[p_{n-1-j}\overline{y}]^{(j)}\\
\sum_{j=0}^{n-2}(-1)^{j}[p_{n-2-j}\overline{y}]^{(j)}\\
\vdots\\
\overline{y}
\end{array}
\right)^{\prime}\\
&=
\left(
\begin{array}{cccc}
&&&\overline{p}_{n}(t)\\
(-1)&&&\overline{p}_{n-1}(t)\\
&\ddots&&\vdots\\
&&(-1)&\overline{p}_{1}(t)
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
\sum_{j=0}^{n-1}(-1)^{j}[p_{n-1-j}\overline{y}]^{(j)}\\
\sum_{j=0}^{n-2}(-1)^{j}[p_{n-2-j}\overline{y}]^{(j)}\\
\vdots\\
\overline{y}
\end{array}
\right),
\end{aligned}\notag
\end{equation}
where we put $p_{0}(t):\equiv1$ for simplicity.
Transforming this into the differential equation, we get
\begin{equation}
(-1)^{n}y^{(n)}(t)+(-1)^{n-1}[\overline{p}_{1}y]^{(n-1)}(t)+\cdots+[\overline{p}_{n-1}y]^{\prime}(t)+\overline{p}_{n}(t)y(t)=0.\notag
\end{equation}
Using the inner product in the first post gives us
\begin{equation}
\left\langle\left(
\begin{array}{c}
x\\
x^{\prime}\\
\vdots\\
x^{(n-1)}
\end{array}
\right),
\left(
\begin{array}{c}
\begin{array}{c}
\sum_{j=0}^{n-1}(-1)^{j}[p_{n-1-j}\overline{y}]^{(j)}\\
\sum_{j=0}^{n-2}(-1)^{j}[p_{n-2-j}\overline{y}]^{(j)}\\
\vdots\\
\overline{y}
\end{array}
\end{array}
\right)\right\rangle
=\sum_{k=0}^{n-1}\sum_{j=0}^{k}(-1)^{j}x^{(n-1-k)}[p_{k-j}\overline{y}]^{(j)}=\text{constant},\label{finaleq}\tag{#}
\end{equation}
which is the desired identity.
I believe that \eqref{finaleq} and (*) in the first post are equivalent.
