[proofplan]
We compare loops based at $x$ with loops based at $y$ by transporting them through the path $\alpha$. A loop $\gamma$ at $x$ is converted into the loop at $y$ obtained by first following $\alpha^{-1}$, then $\gamma$, then $\alpha$. The equivariance of parallel transport under the principal right $G$-action turns the holonomy element $g$ at $p$ into the conjugate $hgh^{-1}$ at $q$. Applying the same argument to the reversed basepoint-change path gives the opposite inclusion.
[/proofplan]
[step:Fix the holonomy convention and the basepoint-change element]
For every piecewise smooth path $\beta: [0,1]\to M$, write
\begin{align*}
\operatorname{PT}_{\beta}: P_{\beta(0)} \to P_{\beta(1)}
\end{align*}
for parallel transport along $\beta$ with respect to $\omega$. We use the standard principal-bundle equivariance of parallel transport: for every $r \in P_{\beta(0)}$ and every $a \in G$,
\begin{align*}
\operatorname{PT}_{\beta}(r a)=\operatorname{PT}_{\beta}(r)a.
\end{align*}
For $r \in P_z$, where $z \in M$, define
\begin{align*}
\operatorname{Hol}_r(\omega):=\{a \in G : \operatorname{PT}_{\delta}(r)=r a \text{ for some piecewise smooth loop } \delta \text{ at } z\}.
\end{align*}
By hypothesis, $h \in G$ is the unique element satisfying $\operatorname{PT}_{\alpha}(p)=q h$. Applying $\operatorname{PT}_{\alpha}^{-1}=\operatorname{PT}_{\alpha^{-1}}$ and equivariance gives
\begin{align*}
\operatorname{PT}_{\alpha^{-1}}(q)=p h^{-1}.
\end{align*}
Indeed, $\operatorname{PT}_{\alpha}(p h^{-1})=\operatorname{PT}_{\alpha}(p)h^{-1}=q$.
[/step]
[step:Send each holonomy element at $p$ to a conjugate holonomy element at $q$]
Let $g \in \operatorname{Hol}_p(\omega)$. By definition of $\operatorname{Hol}_p(\omega)$, there is a piecewise smooth loop $\gamma: [0,1]\to M$ based at $x$ such that
\begin{align*}
\operatorname{PT}_{\gamma}(p)=p g.
\end{align*}
Define the piecewise smooth loop $\beta$ based at $y$ to be the concatenation that first traverses $\alpha^{-1}$, then $\gamma$, then $\alpha$.
Starting from $q$, parallel transport along the three pieces gives
\begin{align*}
\operatorname{PT}_{\alpha^{-1}}(q)=p h^{-1}.
\end{align*}
Using equivariance along $\gamma$,
\begin{align*}
\operatorname{PT}_{\gamma}(p h^{-1})=\operatorname{PT}_{\gamma}(p)h^{-1}=p g h^{-1}.
\end{align*}
Using equivariance along $\alpha$,
\begin{align*}
\operatorname{PT}_{\alpha}(p g h^{-1})=\operatorname{PT}_{\alpha}(p)g h^{-1}=q h g h^{-1}.
\end{align*}
Therefore
\begin{align*}
\operatorname{PT}_{\beta}(q)=q h g h^{-1}.
\end{align*}
Since $\beta$ is a loop based at $y$, this proves $hgh^{-1}\in \operatorname{Hol}_q(\omega)$. As $g$ was arbitrary,
\begin{align*}
h\operatorname{Hol}_p(\omega)h^{-1}\subseteq \operatorname{Hol}_q(\omega).
\end{align*}
[guided]
We want to turn a loop based at $x$ into a loop based at $y$. The path $\alpha$ runs from $x$ to $y$, so a loop at $y$ can first go backward along $\alpha$, then perform a loop at $x$, then return forward along $\alpha$. This is why we define $\beta$ as the concatenation of $\alpha^{-1}$, then $\gamma$, then $\alpha$.
Let $g \in \operatorname{Hol}_p(\omega)$. By the definition of holonomy at $p$, there exists a piecewise smooth loop $\gamma$ at $x$ such that
\begin{align*}
\operatorname{PT}_{\gamma}(p)=p g.
\end{align*}
The loop $\beta$ based at $y$ is constructed so that parallel transport along $\beta$ starts at $q$, moves to the fiber over $x$, applies the known holonomy computation there, and then moves back to the fiber over $y$.
The first transport is along $\alpha^{-1}$. Since $\operatorname{PT}_{\alpha}(p)=q h$, equivariance gives
\begin{align*}
\operatorname{PT}_{\alpha}(p h^{-1})=\operatorname{PT}_{\alpha}(p)h^{-1}=q h h^{-1}=q.
\end{align*}
Thus the inverse transport satisfies
\begin{align*}
\operatorname{PT}_{\alpha^{-1}}(q)=p h^{-1}.
\end{align*}
Now transport this point around $\gamma$. Parallel transport commutes with the right $G$-action, so
\begin{align*}
\operatorname{PT}_{\gamma}(p h^{-1})=\operatorname{PT}_{\gamma}(p)h^{-1}.
\end{align*}
Using the defining equation for $g$ gives
\begin{align*}
\operatorname{PT}_{\gamma}(p h^{-1})=p g h^{-1}.
\end{align*}
Finally, transport forward along $\alpha$. Applying equivariance once more,
\begin{align*}
\operatorname{PT}_{\alpha}(p g h^{-1})=\operatorname{PT}_{\alpha}(p)g h^{-1}.
\end{align*}
Since $\operatorname{PT}_{\alpha}(p)=q h$, this becomes
\begin{align*}
\operatorname{PT}_{\alpha}(p g h^{-1})=q h g h^{-1}.
\end{align*}
Therefore the full loop $\beta$ based at $y$ satisfies
\begin{align*}
\operatorname{PT}_{\beta}(q)=q h g h^{-1}.
\end{align*}
By the definition of $\operatorname{Hol}_q(\omega)$, this says precisely that $hgh^{-1}\in \operatorname{Hol}_q(\omega)$. Since $g$ was an arbitrary element of $\operatorname{Hol}_p(\omega)$, we have proved
\begin{align*}
h\operatorname{Hol}_p(\omega)h^{-1}\subseteq \operatorname{Hol}_q(\omega).
\end{align*}
[/guided]
[/step]
[step:Apply the same argument to the reversed path to get the reverse inclusion]
The element relating $q$ to $p$ along $\alpha^{-1}$ is $h^{-1}$, because
\begin{align*}
\operatorname{PT}_{\alpha^{-1}}(q)=p h^{-1}.
\end{align*}
Applying the previous step with $x$ and $y$ interchanged, with basepoints $q \in P_y$ and $p \in P_x$, and with the path $\alpha^{-1}$ in place of $\alpha$, gives
\begin{align*}
h^{-1}\operatorname{Hol}_q(\omega)h\subseteq \operatorname{Hol}_p(\omega).
\end{align*}
Multiplying this inclusion on the left by $h$ and on the right by $h^{-1}$ yields
\begin{align*}
\operatorname{Hol}_q(\omega)\subseteq h\operatorname{Hol}_p(\omega)h^{-1}.
\end{align*}
Combining the two inclusions,
\begin{align*}
\operatorname{Hol}_q(\omega)=h\operatorname{Hol}_p(\omega)h^{-1}.
\end{align*}
This is the claimed basepoint-change formula for holonomy.
[/step]
