= Solution
For a path $\gamma:[0,1]\to B$, a section $s(t)$ of $\gamma^*E$ is parallel when
$$
\dot s+A(\dot\gamma)s=0.
$$
Existence and uniqueness for this linear ordinary differential equation define the <parallel transport>
$$
\mathcal P_\gamma^{\mathcal A}:E_{\gamma(0)}\longrightarrow E_{\gamma(1)}.
$$
Transport along the reversed path solves the inverse initial-value problem, so
$$
\mathcal P_{\bar\gamma}^{\mathcal A}
=(\mathcal P_\gamma^{\mathcal A})^{-1}.
$$
If $P(t)$ denotes transport from $0$ to $t$, then
$$
\mu(t)=P(t)\mu(0)P(t)^{-1}
$$
satisfies the horizontal equation for the induced endomorphism connection. Uniqueness therefore gives
$$
\mathcal P_\gamma^{\operatorname{End}(\mathcal A)}(\mu)
=\mathcal P_\gamma^{\mathcal A}\,
\mu\,
(\mathcal P_\gamma^{\mathcal A})^{-1}.
$$
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