Without prejudicing the issue on either side, we can follow the Gandalf / Toucan Sam Strategy and follow our nose.
The matter can be addressed by applying the continuity equation to the stress tensor density for the particle:
$$^μ_ν(x) = \int_{-∞}^{+∞} p_ν(s) v^μ(s) δ(x - x(s)) ds, \hspace 1em v^ρ(s) = \frac{dx^ρ}{ds},$$
with the momentum vector being given by:
$$p_ν(s) = m g_{μν}(x(s)) v^μ(s),$$
where $m > 0$ is the mass of the particle, and $s ↦ x(s)$ its worldline with parameterization, $g_{μν} v^μ v^ν = 1$; i.e. $s$ is the proper time, if the convention is chosen of setting $g_{00} = 1$ in the Minkowski limit.
The divergence can be tabulated as follows:
$$\begin{align}
∂_μ^μ_ν(x)
&= \int_{-∞}^{+∞} p_ν(s) v^μ(s) ∂_μ δ(x - x(s)) ds \\
&= \int_{-∞}^{+∞} p_ν(s) \left(-\frac{d}{ds} δ(x - x(s))\right) ds \\
&= -\left[p_ν(s) δ(x - x(s))\right]_{-∞}^{+∞} + \int_{-∞}^{+∞} \frac{dp_ν}{ds} δ(x - x(s)) ds \\
&= \int_{-∞}^{+∞} \frac{dp_ν}{ds} δ(x - x(s)) ds
\end{align}$$
Now ... the continuity equation is for the stress tensor:
$$∇_μT^μ_ν ≡ ∂_μT^μ_ν - Γ_{μν}^ρ T^μ_ρ + Γ_{μρ}^μT^ρ_ν = 0.$$
The Levi-Civita connection is $Γ_{μν}^ρ - C_{μν}^ρ$, where $C_{μν}^ρ$ is the contorsion, or just $Γ-C$ for short. A well-known (to General Relativists) identity for it is given by:
$$(Γ-C)_{μρ}^μ = \frac{1}{2} g^{μν} ∂_ρ g_{μν} = \frac{∂_ρ \sqrt{|g|}}{\sqrt{|g|}}.$$
Applying this to the continuity equation:
$$∂_μT^μ_ν + \frac{∂_ρ \sqrt{|g|}}{\sqrt{|g|}}T^ρ_ν - Γ_{μν}^ρ T^μ_ρ + C_{μρ}^μT^ρ_ν = 0,$$
multiplying by $\sqrt{|g|}$, and noting that the stress tensor density, itself, is given by $^ρ_ν = \sqrt{|g|}T^ρ_ν$, we can write:
we get:
$$∂_μ^μ_ν - Γ_{μν}^ρ^μ_ρ + C_{μρ}^μ^ρ_ν = 0.$$
Upon substitution, we get:
$$
0 = \int_{-∞}^{+∞} \left(\frac{dp_ν}{ds} - Γ_{μν}^ρ(x(s)) p_ρ(s) v^μ(s) + C_{μρ}^μ(x(s)) p_ν(s) v^ρ(s)\right) δ(x - x(s)) ds.
$$
We assume that there is a foliation of the space-time manifold - at least in the vicinity of the worldline - into spacelike 3-surfaces $Σ_s$, such that (1) $x(s) ∈ Σ_s$ for each $s ∈ (-∞,+∞)$ and (ii) we can decompose the delta function into $δ(x_s - x(s')) = δ_{Σ_s}(x_s - x(s))δ(s - s')$. Then an integral of the following form may be worked out as follows:
$$\begin{align}
\int_{Σ_s} \left(\int_{-∞}^{+∞} F(s') δ(x_s - x(s')) ds'\right) dx_s
&= \int_{Σ_s} \left(\int_{-∞}^{+∞} F(s') δ_{Σ_s}(x_s - x(s))δ(s - s') ds'\right) dx_s \\
&= \int_{-∞}^{+∞} F(s') \left(\int_{Σ_s} δ_{Σ_s}(x_s - x(s)) dx_s\right) δ(s - s') ds' \\
&= \int_{-∞}^{+∞} F(s') δ(s - s') ds' \\
&= F(s).
\end{align}$$
Applying this to the reworked continuity equation, we get:
$$\frac{dp_ν}{ds} - Γ_{μν}^ρ p_ρ v^μ + C_{μρ}^μ p_ν v^ρ = 0,\label{1}\tag{1}$$
where all $x$-dependent quantities are evaluated on the worldline at $x(s)$. The first two terms are the momentum-space version of the terms that appear in the geodesic and autoparallel equations:
$$\frac{dp_ν}{ds} - Γ_{μν}^ρ p_ρ v^μ = m g_{μν} \left(\frac{dv^μ}{ds} + Γ_{νρ}^μ v^ν v^ρ\right). \label{2}\tag{2}$$
This requires on that the connection $Γ$ be metrical and involves the chain rule on $g_{μν}(x(s))$ in $p_ν = g_{μν}v^μ$, along with the metrical property of the connection:
$$
\frac{dp_ν}{ds} = m\left(\frac{dg_{μν}}{ds} v^μ + g_{μν} \frac{dv^μ}{ds}\right), \hspace 1em
\frac{dg_{μν}}{ds} = ∂_ρ g_{μν} v^ρ = \left(g_{σν} Γ_{ρμ}^σ + g_{μσ} Γ_{ρν}^σ\right) v^ρ \\
⇒ \frac{dp_ν}{ds} = m\left(g_{σν} Γ_{ρμ}^σ + g_{μσ} Γ_{ρν}^σ\right) v^ρ v^μ + m g_{μν} \frac{dv^μ}{ds}.
$$
We can rewrite
$$Γ_{μν}^ρ p_ρ v^μ = Γ_{ρν}^σ p_σ v^ρ = m g_{μσ} Γ_{ρν}^σ v^μ v^ρ.$$
Therefore, upon cancellation, we get:
$$\begin{align}
\frac{dp_ν}{ds} - Γ_{μν}^ρ p_ρ v^μ
&= m g_{σν} Γ_{ρμ}^σ v^ρ v^μ + m g_{μν} \frac{dv^μ}{ds} \\
&= m g_{μν} \frac{dv^μ}{ds} + m g_{μν} Γ_{ρσ}^μ v^ρ v^σ \\
&= m g_{μν} \left(\frac{dv^μ}{ds} + Γ_{ρσ}^μ v^ρ v^σ\right).
\end{align}$$
Rewriting the contorsion term in ($\ref{1}$),
$$C_{μρ}^μ p_ν v^ρ = -p_ν N = -m g_{μν} N v^μ, \hspace 1em N = -C_{σρ}^σ v^ρ,$$
divide out the mass $m > 0$:
$$
g_{μν} \left(\frac{dv^μ}{ds} + Γ_{ρσ}^μ v^ρ v^σ - N v^μ\right) = 0
$$
and invert the metric in $g_{μν}$ to get:
$$\frac{dx^μ}{ds} = v^μ, \hspace 1em \frac{dv^μ}{ds} + Γ_{ρσ}^μ v^ρ v^σ = N v^μ. \tag{3}$$
That's a variation of the auto-parallel equation, but with the need to reparametrize $s$. If you reset the parameter to $S$, such that
$$\frac{d}{ds}ln\left(\frac{dS}{ds}\right) = N = -C_{σρ}^σ v^ρ, \tag{4}$$
then you'll get:
$$\frac{dx^μ}{dS} = v^μ, \hspace 1em \frac{dv^μ}{dS} + Γ_{ρσ}^μ v^ρ v^σ = 0. \tag{5}$$
There's a warping of some sort on the proper time $s$ into $S$.
The contorsion is warping the time.
I did these calculations extemporaneously. You should check it closely. It needs a once-over. I might try it again at a later time. I'm pretty sure of ($\ref{1}$) and ($\ref{2}$) is fairly standard and well-known.
