[proofplan] Fix a point in an overlap and compare the two ways of writing the same point of the fiber using the two systems of local sections. Substituting the relations $s'_i = s_i a_i$ and $s'_j = s_j a_j$ into the defining equation for $g'_{ij}$ gives an equality of two right translates of $s_i(x)$. The freeness of the principal right action then identifies the two group elements, and a final multiplication in $G$ gives the claimed conjugation formula. [/proofplan] [step:Fix an overlap point and write both transition identities there] Fix indices $i,j \in I$ and a point $x \in U_i \cap U_j$. The transition functions are defined by \begin{align*} s_j(x) = s_i(x)g_{ij}(x) \end{align*} and \begin{align*} s'_j(x) = s'_i(x)g'_{ij}(x). \end{align*} The change of local section is given by \begin{align*} s'_i(x) = s_i(x)a_i(x) \end{align*} and \begin{align*} s'_j(x) = s_j(x)a_j(x). \end{align*} [/step] [step:Substitute the changed sections into the defining relation for $g'_{ij}$] Using $s'_j(x) = s_j(x)a_j(x)$ and then $s_j(x) = s_i(x)g_{ij}(x)$, we obtain \begin{align*} s'_j(x) = s_i(x)g_{ij}(x)a_j(x). \end{align*} Using $s'_i(x) = s_i(x)a_i(x)$ in the defining relation $s'_j(x) = s'_i(x)g'_{ij}(x)$ gives \begin{align*} s'_j(x) = s_i(x)a_i(x)g'_{ij}(x). \end{align*} Therefore \begin{align*} s_i(x)g_{ij}(x)a_j(x) = s_i(x)a_i(x)g'_{ij}(x). \end{align*} [guided] We compare the same point $s'_j(x)$ in the fiber $\pi^{-1}(x)$ in two different ways. First, the changed section $s'_j$ is related to $s_j$ by the map $a_j$, so \begin{align*} s'_j(x) = s_j(x)a_j(x). \end{align*} The original transition function $g_{ij}$ satisfies $s_j(x) = s_i(x)g_{ij}(x)$, hence substitution gives \begin{align*} s'_j(x) = s_i(x)g_{ij}(x)a_j(x). \end{align*} Second, the primed transition function $g'_{ij}$ is defined by \begin{align*} s'_j(x) = s'_i(x)g'_{ij}(x). \end{align*} The changed section $s'_i$ satisfies $s'_i(x) = s_i(x)a_i(x)$, so this becomes \begin{align*} s'_j(x) = s_i(x)a_i(x)g'_{ij}(x). \end{align*} Both expressions are equal to the same point $s'_j(x) \in \pi^{-1}(x)$. Thus \begin{align*} s_i(x)g_{ij}(x)a_j(x) = s_i(x)a_i(x)g'_{ij}(x). \end{align*} This equality is the point at which the principal action enters: both sides are right translates of the single point $s_i(x)$ by elements of $G$. [/guided] [/step] [step:Use freeness of the principal right action to identify the group elements] Define $p := s_i(x) \in \pi^{-1}(x)$, define $h := g_{ij}(x)a_j(x) \in G$, and define $k := a_i(x)g'_{ij}(x) \in G$. The preceding equality says \begin{align*} ph = pk. \end{align*} Since $P$ is a principal right $G$-bundle, the right action of $G$ on each fiber is free. Therefore $h = k$, which gives \begin{align*} g_{ij}(x)a_j(x) = a_i(x)g'_{ij}(x). \end{align*} [/step] [step:Multiply in $G$ to obtain the transformed transition function] Multiplying the equality \begin{align*} g_{ij}(x)a_j(x) = a_i(x)g'_{ij}(x) \end{align*} on the left by $a_i(x)^{-1}$ in the group $G$, we obtain \begin{align*} a_i(x)^{-1}g_{ij}(x)a_j(x) = g'_{ij}(x). \end{align*} Equivalently, \begin{align*} g'_{ij}(x) = a_i(x)^{-1}g_{ij}(x)a_j(x). \end{align*} Because $i,j \in I$ and $x \in U_i \cap U_j$ were arbitrary, the formula holds on every overlap $U_i \cap U_j$. [/step]