A connection is metric compatible iff parallel transport is isometric

Assume compatibility. Along a frame segment, let $v,w$ be parallel coefficient columns. The chain rule gives $(H\circ \gamma {)}^{\mathrm{\prime }}=B^{T}H+HB$ by [F1]. Then $(v^{T}Hw{)}^{\mathrm{\prime }}=(-Bv{)}^{T}Hw+v^T(B^{T}H+HB)w+v^{T}H(-Bw)=0$. Their inner product is constant on each smooth piece and, by continuity, across corners. Hence endpoint transport preserves $h$ for all pairs, and is an isometry by its linear invertibility.

Conversely assume all transports are isometries. Fix one frame near $p$ and a curve through $p$ with tangent $z$. Choose any initial coefficient vectors $v_0,w_0$ and their parallel solutions. Isometry makes $v^T(H\circ \gamma )w$ constant. Differentiating at the initial time and using [F3] gives $v_0^T(zH-\omega (z{)}^{T}H-H\omega (z))w_0=0$. Testing on the finitely many pairs of standard basis vectors makes this matrix zero. At an interior point every coordinate direction is realized by a short coordinate line. At a boundary point use coordinate lines within the boundary for tangential basis directions and the inward one-sided normal line for the last direction; the one-sided derivative gives the same identity. Linearity then covers every tangent vector, including outward ones, without claiming an outward curve lies in the manifold.

The matrix identity from step 1.2 is the compatibility criterion [F1]. The empty manifold and zero-rank bundle satisfy both conditions vacuously; in dimension zero compatibility has no nonzero directional test and all curves are constant. Rank one is the same one-entry calculation. The converse only chooses finitely many initial vectors at a fixed point, so neither direction invokes AC.