Let me try to tackle the LPO case. I will even try to show that it doesn't matter whether we assume our Cauchy sequences to have a modulus or not. (Please check me carefully because I know I've made mistakes on similar questions before.)
The easy direction is to prove that LPO for Cauchy reals (with or without modulus) implies LPO for natural numbers. To this effect, let $b\colon\mathbb{N}\to\{0,1\}$ and we wish to show that either $\exists n.(b(n)=1)$ or that $\forall n.(b(n)=0)$. Define $u_n \in \mathbb{Q}$ as $\sum_{k=0}^{n} 2^{-k}\,b(k)$. Clearly this is Cauchy-with-explicit-modulus, so it defines a real $x$ (viꝫ. the limit of $(u_n)$): by LPO on the Cauchy reals (with or a fortiori without modulus), we have either $x\mathrel{\#}0$ (“is apart from”) or $x=0$. In the former case, $x>2^{-k}$ for some $k$ so it is easy to see that some $b(n)=1$, and in the latter case, clearly $b(n)=0$ for all $n$. ∎
Now for the other direction. We assume LPO for natural numbers. Let $x$ be a Cauchy real, defined as the limit of a sequence $(u_n)$ of rationals, which we do not assume to have a modulus so as to have the most general result. We want to show that $x\mathrel{\#}0$ or that $x=0$. For notational convenience, we assume w.l.o.g. (replacing $u_n$ by $|u_n|$ and $x$ by $|x|$) that $u_n\geq 0$ and $x\geq 0$, so now we want to show that $x>0$ or that $x=0$.
Let $k,n\in\mathbb{N}$. By LPO for the natural numbers (and because comparisons of rationals are decidable), either $u_m > 2^{-k}$ for some $m\geq n$ or $u_m \leq 2^{-k}$ for all $m\geq n$. Since these two cases are exclusive, we can define $b(k,n)=1$ in the former and $b(k,n)=0$ in the latter.
Now let $k\in\mathbb{N}$ (but allow $n$ to vary). By a second application of LPO on the natural numbers, either there is $n$ such that $b(k,n)=0$ or we have $b(k,n)=1$ for all $n$. Again, these two cases are exclusive, so we can define $b(k)=0$ in the former and $b(k)=1$ in the latter.
Now by a third application of LPO on the natural numbers, either there is $k$ such that $b(k)=1$ or we have $b(k)=0$ for all $k$.
This shows that either $\exists k. \forall n. \exists m\geq n. (u_m > 2^{-k})$ or $\forall k. \exists n. \forall m\geq n. (u_m \leq 2^{-k})$. In the former case we have $x\geq 2^{-k}$ so $x>0$. In the latter case, we have $x=0$. ∎
PS [added in edit]: in the above, I wanted to show that LPO for natural numbers implies LPO for Cauchy reals even without modulus; but since the question was about Cauchy reals with modulus, I should point out that the proof (of the second part) simplifies: if we assume $|u_p-u_q|\leq 2^{-n}$ when $p,q\geq n$, then $x>0$ is equivalent to $\exists n.(u_n>2^{-n})$ and $x=0$ to $\forall n.(u_n\leq 2^{-n})$, so we need to show that either the former or the latter holds, and that is a single application of LPO for $\mathbb{N}$.