When does the Lie Derivative coincides with the Covariant Derivative?

Let (M,g) be a Riemannian manifold and $$\nabla$$ a connection on $$TM$$. Let $$X \in \mathfrak{X}(M)$$ be a smooth vector field on M, and denote by $$\theta$$ its flow. Let $$Y$$ be a smooth vector field on M that is smooth along the flow of $$X$$ at $$p$$. Denote by $$D_t$$ the induced covariant derivative along the flow of $$X$$ at $$p$$. Under what condition(s) on $$\nabla$$ can we have:

$$L_X Y = D_t Y$$

Recall that :
$$L_X Y = \underset{t \rightarrow 0}{lim} \frac{d(\theta_{-t})_{\theta^{p}(t)}(Y_{\theta^{p}(t)})-Y_{p}}{t} = [X,Y]$$

Thank you !

I'm a bit confused why you take $$(M,g)$$ to be Riemannian but then separately impose a connection $$\nabla$$, which is not necessarily the Levi-Civita connection (so the metric structure is not used). Also, $$\nabla$$ is usually used for the covariant derivative, so what you call $$D_t$$ seems to be the same thing as $$\nabla_X$$?
I think what you're asking is when does $$\nabla_XY = \mathcal{L}_XY$$ for all $$X, Y\in\mathfrak{X}(M)$$? If so, never: $$\nabla$$ needs to satisfy $$\nabla_{fX}Y = f\nabla_XY$$ for all $$f\in C^\infty(M)$$. By contrast, $$\mathcal{L}_{fX}Y = [fX,Y] = f\mathcal{L}_XY - (Yf)X,$$ and for non-zero $$Y$$ you can always choose $$f$$ so that $$Yf$$ doesn't vanish.