[proofplan]
We prove the equality by comparing parallel transport from the two points $p$ and $p h$ in the same fibre. The key structural fact is that a principal connection is equivariant under the right $G$-action, so parallel transport commutes with right multiplication: transporting $p h$ along a loop gives the transport of $p$, then multiplied by $h$. This converts a holonomy element $g$ based at $p$ into the conjugate $h^{-1}g h$ based at $p h$, and applying the same argument with $h^{-1}$ gives the reverse inclusion.
[/proofplan]
[step:Use equivariance of the connection to compare parallel transports from $p$ and $p h$]
Let $R_h:P\to P$ denote the right translation map defined by $R_h(q)=q h$. Since $\omega$ is a principal connection, the horizontal distribution $\ker\omega\subset TP$ is invariant under right translation in the following sense: if $v\in T_qP$ satisfies $\omega_q(v)=0$, then $(dR_h)_q(v)\in T_{q h}P$ satisfies $\omega_{q h}((dR_h)_q(v))=0$.
Let $\gamma:[0,1]\to M$ be a piecewise smooth path with $\gamma(0)=x$. Let
\begin{align*}
\widetilde{\gamma}_p:[0,1]\to P
\end{align*}
be the $\omega$-horizontal lift of $\gamma$ with $\widetilde{\gamma}_p(0)=p$. Then $R_h\circ\widetilde{\gamma}_p:[0,1]\to P$ is an $\omega$-horizontal lift of $\gamma$ with initial value
\begin{align*}
(R_h\circ\widetilde{\gamma}_p)(0)=p h.
\end{align*}
By uniqueness of horizontal lifts, this curve is the horizontal lift of $\gamma$ starting at $p h$. Evaluating at $t=1$ gives
\begin{align*}
\operatorname{PT}^{\omega}_{\gamma}(p h)=\operatorname{PT}^{\omega}_{\gamma}(p)h.
\end{align*}
[guided]
The point of this step is to justify the only geometric input used later: parallel transport is compatible with the principal right action. Define the right translation map $R_h:P\to P$ by the rule
\begin{align*}
R_h(q)=q h.
\end{align*}
Because $\omega$ is a principal connection, right translation preserves horizontal tangent vectors. Explicitly, if $v\in T_qP$ is horizontal, meaning $\omega_q(v)=0$, then $(dR_h)_q(v)\in T_{q h}P$ is also horizontal, meaning
\begin{align*}
\omega_{q h}((dR_h)_q(v))=0.
\end{align*}
Now let $\gamma:[0,1]\to M$ be a piecewise smooth path with $\gamma(0)=x$, and let
\begin{align*}
\widetilde{\gamma}_p:[0,1]\to P
\end{align*}
be the $\omega$-horizontal lift of $\gamma$ beginning at $p$. Thus $\pi(\widetilde{\gamma}_p(t))=\gamma(t)$ for every $t\in[0,1]$, $\widetilde{\gamma}_p(0)=p$, and the velocity of $\widetilde{\gamma}_p$ is horizontal wherever it is defined.
Consider the translated curve $R_h\circ\widetilde{\gamma}_p:[0,1]\to P$. Since the right action is fibre-preserving, we have
\begin{align*}
\pi((R_h\circ\widetilde{\gamma}_p)(t))=\pi(\widetilde{\gamma}_p(t)h)=\pi(\widetilde{\gamma}_p(t))=\gamma(t).
\end{align*}
So $R_h\circ\widetilde{\gamma}_p$ still lies over $\gamma$. Its initial value is
\begin{align*}
(R_h\circ\widetilde{\gamma}_p)(0)=\widetilde{\gamma}_p(0)h=p h.
\end{align*}
Finally, its velocity is horizontal because right translation preserves the horizontal distribution. Therefore $R_h\circ\widetilde{\gamma}_p$ is the $\omega$-horizontal lift of $\gamma$ starting at $p h$. By uniqueness of horizontal lifts, its endpoint is the parallel transport endpoint from $p h$, and hence
\begin{align*}
\operatorname{PT}^{\omega}_{\gamma}(p h)=(R_h\circ\widetilde{\gamma}_p)(1)=\widetilde{\gamma}_p(1)h=\operatorname{PT}^{\omega}_{\gamma}(p)h.
\end{align*}
[/guided]
[/step]
[step:Show that every holonomy element at $p$ gives a conjugate holonomy element at $p h$]
Let $g\in\operatorname{Hol}_p(\omega)$. By definition of $\operatorname{Hol}_p(\omega)$, there exists a piecewise smooth loop $\gamma:[0,1]\to M$ based at $x$, meaning $\gamma(0)=\gamma(1)=x$, such that
\begin{align*}
\operatorname{PT}^{\omega}_{\gamma}(p)=p g.
\end{align*}
Using the equivariance identity from the previous step for this loop $\gamma$, we obtain
\begin{align*}
\operatorname{PT}^{\omega}_{\gamma}(p h)=\operatorname{PT}^{\omega}_{\gamma}(p)h=(p g)h.
\end{align*}
Associativity of the right action gives
\begin{align*}
(p g)h=p(gh)=(p h)(h^{-1}g h).
\end{align*}
Therefore the same loop $\gamma$ realizes $h^{-1}g h$ as a holonomy element at $p h$, so
\begin{align*}
h^{-1}g h\in\operatorname{Hol}_{p h}(\omega).
\end{align*}
Since $g\in\operatorname{Hol}_p(\omega)$ was arbitrary,
\begin{align*}
h^{-1}\operatorname{Hol}_p(\omega)h\subset \operatorname{Hol}_{p h}(\omega).
\end{align*}
[/step]
[step:Apply the same argument with $h^{-1}$ to obtain the reverse inclusion]
Apply the previous inclusion to the point $p h\in P_x$ and the group element $h^{-1}\in G$. This gives
\begin{align*}
h\operatorname{Hol}_{p h}(\omega)h^{-1}\subset \operatorname{Hol}_{(p h)h^{-1}}(\omega).
\end{align*}
Since $(p h)h^{-1}=p$, this becomes
\begin{align*}
h\operatorname{Hol}_{p h}(\omega)h^{-1}\subset \operatorname{Hol}_p(\omega).
\end{align*}
Multiplying this subset relation on the left by $h^{-1}$ and on the right by $h$ gives
\begin{align*}
\operatorname{Hol}_{p h}(\omega)\subset h^{-1}\operatorname{Hol}_p(\omega)h.
\end{align*}
Together with the inclusion proved in the previous step, we conclude
\begin{align*}
\operatorname{Hol}_{p h}(\omega)=h^{-1}\operatorname{Hol}_p(\omega)h.
\end{align*}
This is the desired conjugacy relation for the holonomy groups based at $p$ and $p h$.
[/step]
