Are there any known commutation rules between the $X$ gate and the $CH$ gate?
2 Answers
There are two cases: $X$ on the controlled qubit and $X$ on the target.
$X$ on the controlled qubit
For the first case, note that for any controlled unitary $CU$, we have
$$ (X_1\otimes I_2)\circ C_1U_2 = C_1U_2^\dagger\circ (X_1\otimes U_2)\tag1 $$
which is easy to prove by separately considering the action of both sides on $|0\rangle|\psi\rangle$ and $|1\rangle|\psi\rangle$.
$X$ on the target qubit
For the second case, we can use $XH=HZ$ to write
$$ \begin{align} (I_1\otimes X_2)\circ C_1H_2 &= \begin{bmatrix}X&\\& X\end{bmatrix} \begin{bmatrix}I&\\& H\end{bmatrix}\\ &= \begin{bmatrix}X&\\& XH\end{bmatrix}\\ &= \begin{bmatrix}X&\\& HZ\end{bmatrix}\\ &= \begin{bmatrix}I&\\& H\end{bmatrix} \begin{bmatrix}X&\\& I\end{bmatrix} \begin{bmatrix}I&\\& Z\end{bmatrix}\\ &= C_1H_2\circ\begin{bmatrix}X&\\& I\end{bmatrix}\circ C_1Z_2 \end{align}\tag2 $$
where the matrix $\mathrm{diag}(X, I)$ represents the quantum gate that flips the second qubit if the first qubit is in the $|0\rangle$ state. We can synthesize this gate as
$$ \begin{align} \begin{bmatrix}X&\\& I\end{bmatrix} &= \begin{bmatrix}&I\\I&\end{bmatrix} \begin{bmatrix}I&\\& X\end{bmatrix} \begin{bmatrix}&I\\I&\end{bmatrix}\\ &= (X_1\otimes I_2)\circ C_1X_2\circ(X_1\otimes I_2). \end{align}\tag3 $$
Substituting $(3)$ into $(2)$, we end up with
$$ (I_1\otimes X_2)\circ C_1H_2 = C_1H_2\circ(X_1\otimes I_2)\circ C_1X_2\circ(X_1\otimes I_2)\circ C_1Z_2.\tag4 $$
Reducing the number of controlled gates
The technique above allows us to reduce the number of resulting controlled gates by exploiting identities such as $XZ=-iY$
$$ \begin{align} (I_1\otimes X_2)\circ C_1H_2 &= \begin{bmatrix}X&\\& X\end{bmatrix} \begin{bmatrix}I&\\& H\end{bmatrix}\\ &= \begin{bmatrix}X&\\& XH\end{bmatrix}\\ &= \begin{bmatrix}X&\\& HZ\end{bmatrix}\\ &= \begin{bmatrix}I&\\& H\end{bmatrix} \begin{bmatrix}X&\\& Z\end{bmatrix}\\ &= C_1H_2\circ\begin{bmatrix}X&\\& X\end{bmatrix} \begin{bmatrix}I&\\& XZ\end{bmatrix}\\ &= C_1H_2\circ(I_1\otimes X_2)\circ\begin{bmatrix}I&\\& -iY\end{bmatrix}\\ &= C_1H_2\circ(I_1\otimes X_2)\circ S_1^\dagger \circ C_1Y_2 \end{align}\tag5 $$
which agrees with the identity obtained by Craig Gidney using quirk.
With a few minutes in Quirk you can confirm that
$$X_c \cdot \text{CH}_{c \rightarrow t} = \text{CH}_{c \rightarrow t} \cdot X_c H_t$$
and
$$X_t \cdot \text{CH}_{c \rightarrow t} = \text{CH}_{c \rightarrow t} \cdot X_t \text{CY}_{c \rightarrow t} S^\dagger_c$$