Linear control theory and transition matrix problem So I want to prove the following statement:

Consider an $n$-dimensional time-varying system $\dot{x} = A(t)x$, where $A(t)$ is continuous. If $A^{\intercal}(t)=-A(t)$ $\forall t\in\mathbb{R}$, then $\Phi^{\intercal}(t,s)=\Phi^{-1}(t,s)$, where $\Phi(t,s)$ is the state transition matrix.

So first of all, I know the following:
\begin{align}
\frac{\partial}{\partial s}\Phi(t,s) = -\Phi(t,s)A(s) \iff \frac{\partial}{\partial s}\Phi^{\intercal}(t,s) = -A^{\intercal}(s)\Phi^{\intercal}(t,s) = A(s)\Phi^{\intercal}(t,s).
\end{align}
So how do I go from here to show $\Phi^{\intercal}(t,s)=\Phi^{-1}(t,s)$? It feels like I am on the track, but I am stuck.
Edit
If you are unfamiliar with the transition matrix, this appears in the solution to system of differential equations:
$x(t)=\Phi(t,s)x(s)$
where $x(s)$ is the initial state.
 A: Consider
$$\dot{x} = Ax, \; x(t_0) = x_0 \tag{1}\label{1}$$
$$\dot{z} = -A^\top z, \; z(t_0) = z_0 \tag{2}\label{2}$$
Let $\Phi_x(t,s)$ be the state transition matrix for $(1)$, and $\Phi_z(t,s)$ be the transition matrix for $(2)$. These satisfy the differential equations
\begin{cases}
    \frac{\partial}{\partial t} \Phi_x(t,s) = A \Phi_x(t,s),& \Phi_x(s,s) = I\\
    \frac{\partial}{\partial t} \Phi_z(t,s) = -A^\top \Phi_z(t,s),& \Phi_z(s,s) = I
\end{cases}
Now consider
$$\frac{\partial}{\partial t}\Phi_z^\top(t,s) \Phi_x(t,s) = (\frac{\partial}{\partial t}\Phi_z^\top(t,s)) \Phi_x(t,s) + \Phi_z^\top(t,s) (\frac{\partial}{\partial t} \Phi_x(t,s)) = -\Phi_z^\top(t,s) A \Phi_x(t,s) + \Phi_z^\top(t,s) A \Phi_x(t,s) = 0$$
First, we conclude that $\Phi_z^\top(t,s) \Phi_x(t,s)$ is constant. Second, note that $\Phi_z^\top(s,s) \Phi_x(s,s) = I$. These statements imply that $\Phi_z^\top(t,s) \Phi_x(t,s) = I$ for all $t$. We can then rearrange the equation to obtain $\Phi_z^\top(t,s) = \Phi_x^{-1}(t,s)$. In the special case when $A = -A^\top$, we can see that
$$\Phi^\top (t,s) = \Phi^{-1}(t,s).$$
A: Hint: It suffices to note that $\|\Phi(t,s)x\| = \|x\|$ for every initial state vector $x$. Try to show that
$$
\frac {\partial }{\partial t} \|\Phi(t,s) x\|^2 = 0.
$$

In detail: a matrix $M$ satisfies $M^\top = M^{-1}$ if and only if it satisfies $\|Mx\| = \|x\|$ for all vectors $x$. So, it suffices to show that we have $\|\Phi(t,s)x\| = \|x\|$ for every possible $x$.
Note that for any $x$, $x(t) = \Phi(t,s)x$ satisfies the differential equation $\dot x(t) = A(t) x$. It follows that
\begin{align}
\frac \partial{\partial t} \|\Phi(t,s)x\|^2 &= 
\frac \partial{\partial t} [x(t)^\top x(t)] = \dot x^\top(t) x(t) + x^\top(t) \dot x(t)
\\ & = 
[A(t) x(t)]^\top x(t) + x^\top(t)[A(t) x(t)]
\\ & =
x^\top(t)A^\top(t)x(t) + x^\top(t)A(t)x(t)
\\ & = x^\top(t)[-A(t)]x(t) + x^\top A(t) x(t) = 0.
\end{align}
Thus, it must hold that $\|\Phi(t,s)x\|$ is constant. It follows that
$$
\|\Phi(t,s) x\| = \|\Phi(s,s)x\| = \|Ix\| = \|x\|.
$$
That is, for all $t$, the matrix $M = \Phi(t,s)$ satisfies $\|Mx\| = \|x\|$ for all $x$. Thus, for all $t$, the matrix $M = \Phi(t,s)$ satisfies $M^\top = M^{-1}$, which is what we wanted.
