$SO(p,q)$ Fundamental Weights? The weights in the $D^{n-1}$ and $D^{n}$ spinor representations of $SO(2n)$ are of the form
$$\frac{1}{2}(\pm e_1 \pm e_2 \pm ... \pm e_{n-1} \pm e_n)$$
such that the products of all the $\pm 1$'s are either $+1$ or $-1$ to match the signs of the highest weights
$$\mu_{n-1} = \frac{1}{2}(+e_1 +e_2 +... + e_{n-1} + e_n)  \ , \ \ \mu_{n} = \frac{1}{2}(+e_1 +e_2 +... + e_{n-1} - e_n).$$
This offers a way to see that $D^{n-1}$ and $D^{n}$ are complex conjugates when $n$ is odd (by looking at $-\mu_{n-1}$ and $-\mu_n$), for example (ignoring that $n>2$ should hold) in $\mathrm{SO}(4) = \mathrm{SO}(2 \cdot 2)$ they are not complex conjugates.
Question: Is there a (simple/easy) way to modify this fundamental weight analysis to the case of the Lorentz group $\mathrm{SO}(1,3)$ (and more generally $\mathrm{SO}(p,q)$)?
I would imagine it amounts to ending up with highest weights as above but with $-e_1$ instead of $+e_1$ in each,
$$\tilde{\mu}_{n-1} = \frac{1}{2}(-e_1 +e_2 +... + e_{n-1} + e_n)  \ , \ \ \tilde{\mu}_{n} = \frac{1}{2}(-e_1 +e_2 +... + e_{n-1} - e_n)$$
so that now they are complex conjugates when $n$ is even instead of odd, e.g. they are complex conjugates for $\mathrm{SO}(1,2 \cdot 2-1) = \mathrm{SO}(1,3)$.
The end result is usually found in terms of Clifford algebras, but if you're allowed to do something like re-define the Weyl reflections, or re-define the fundamental weights so they satisfy $2(\alpha_{\mu},\mu_{\nu})/(\alpha_{\mu},\alpha_{\mu}) = 2 \eta_{\mu \nu}$ so that we end up with $-e_1$, without something else going wrong?
(It seems that something like this discussion of 'reality structures' may help with justifying this, whatever they are and however they relate to things like the fundamental weights, inner products etc...)
 A: I'll talk about Lie algebras because I feel more comfortable with them than with groups. I am quite sure though that translating what I write to the group setting would need at most minor adjustments, and some of the references below actually talk about groups already. --
First, it's important to note that the representations we deal with (and by the way, I think the two you are interested in are usually called the two "half-spin" representations), or any highest weight representations, a priori are complex representations of the complexified Lie algebras, i.e. you are a priori describing representations of $\mathfrak{so}(2n) \otimes {\mathbb C}$.
But surely what one would be interested in are representations of the real Lie algebras, in this case, $\mathfrak{so}_{p,q}$ with $p+q=2n$ (in the following conventionally $p \le q$). In principle this is easy: All these sit inside the complex Lie algebra $\mathfrak{so}(2n) \otimes {\mathbb C} \simeq \mathfrak{so}_{p,q} \otimes \mathbb C$ (the complexifications are all isomorphic, Complexification of $\mathfrak{so}(p,q)$), and you just get the representation by restricting the action of $\mathfrak{so}(2n) \otimes {\mathbb C}$ to the real form of your choice $\mathfrak{so}_{p,q}$. Confer my lengthy answer here.
In that answer I point out that the trickier thing to see in this restriction business is what complex conjugation does, i.e. which representations of a given real form are conjugate, and that that very much depends on which real form we're looking at. You describe what happens for the compact form $\mathfrak{so}_{0,2n}$: Indeed complex conjugation flips the weights $\mu_{n-1} \leftrightarrow \mu_{n}$, meaning that the two half-spin representations (more precisely: the restrictions of the half-spin representations to this compact form), being the ones which have those as highest weights, are conjugates of each other. Now what changes if we look at general $\mathfrak{so}_{p,q}$? Your attempt is on a wrong track: For each such form, the restrictions of the half-spin representations come from the very same complex half-spin representations with highest weights $\mu_{n-1}$ and $\mu_n$. Nothing changes about the weights themselves: What changes is how complex conjugation acts on the weights.
Namely, in the case of compact forms, complex conjugation always acts on the root and weight lattice as $-id$; but in the case of other forms, it does not.
So now we have to check how complex conjugation operates on the root lattice and weight lattice for each real form $\mathfrak{so}_{p,q}$. See the general philosophy here: https://math.stackexchange.com/a/3298058/96384, which very much summarises my very poor understanding of Tits' article Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque.
Turns out the Satake diagram has an arrow between the two "horns" of the Dynkin diagram, (which will make the restrictions of the two half-spin representations to the real Lie algebra conjugate to each other), iff $q-p (=2n-2p) \equiv 2$ mod $4$. Indeed, we have the following table:

(One way to see this: For $q \ge p+4$ the Lie algebra $\mathfrak{so}_{p,q}$ has anisotropic kernel $\mathfrak{so}_{0, 2n-2p}$ i.e. this compact subalgebra, and that one is dealt with in the classical compact case as in your question. In the remaining cases $q=p$ we have the split form without arrow, and $q=p+2$ gives the quasi-split form with an arrow. -- The visualisations in Onishchik/Vinberg as well as by Tits in the Boulder Proceedings (p.56/57) are unfortunate insofar as they do not draw arrows on black vertices, hence seem to not go into this case distinction for what they call $\mathfrak{so}_{p,2l-p}$; but they have the two remaining cases right. See my thesis pp. 86/87 for a more elaborate discussion.)
This matches that in the Wikipedia article for $q-p \equiv 2, 6$ mod $8$ we find "truly complex" representations, i.e. not equivalent to their own conjugate.
In particular, the two half-spin representations are conjugates of each other (hence not equivalent to their conjugates) for $\mathfrak{so}_{1,3}$ but not for $\mathfrak{so}_{0,4}$. And this is not because anything about their defining weights $\mu_{1}, \mu_2$ would change, but because when you complexify $\mathfrak{so}_{1,3}$ and look at how complex conjugation acts on roots, it switches two basis roots (which by extending the action to the weight lattice switches $\mu_2 \leftrightarrow \mu_{1}$), whereas if you complexify $\mathfrak{so}_{0,4}$ and look at how complex conjugation acts on roots, it sends each root to its own negative, which by extending the action to the weight lattice sends $\mu_2 \mapsto -\mu_2$ which, as you have noticed, gives a representation equivalent to $\mu_2$, and $\mu_{1} \mapsto -\mu_{1}$ whose highest weight irrep is up to equivalence the one of $\mu_1$ again.

If interested, it remains to discuss the cases $q-p \equiv 0,4$ mod $8$, where the half-spin reps are equivalent to their own conjugate. If that happens, there is a further distinction between quaternionic (for physicists: "pseudoreal") and real representations, which is a bit intricate. Compare https://mathoverflow.net/q/323969/27465. Actually, a generalisation of this to more general fields than $\mathbb R$ is what Tits' article Représentations linéaires irréductibles ... quoted above is mostly about; the main result about our case is the first proposition in part 6 (p. 212) there. Before that (p. 211) he reduces to the compact case, and for the compact case, another source is Bourbaki's Lie Algebras and Groups, ch. IX §7 no.2 proposition 1, as quoted in my answer to What property of the root system means a Lie algebra has complex structure?.
Eventually, the method in Bourbaki is different from the one in Tits's article, but in both of them the criterion for such a representation to be quaternionic or real is the parity of a certain invariant; I don't see right now why in general the two methods would give the same invariant (never mind: equality of those invariants is shown in Bourbaki Lie Groups and Algebras ch. VI §1 no. 10 prop. 29 and corollary), but in our case I can compute that both of them, for both the half-spin representations of highest weights $\mu_n$ and $\mu_{n-1}$, are given by
$$\frac12 (n^2-n)$$
which is odd for $n \equiv 2,3$ and even for $n \equiv 0,1$ mod $4$. Since we're already in the case of even $n$, our dichotomy is

both half-spin rep. of $\mathfrak{so}_{0,2n}$ real $\Leftrightarrow$ $n \equiv 0$ mod $4$
both half-spin rep. of $\mathfrak{so}_{0,2n}$ quaternionic $\Leftrightarrow$ $n \equiv 2$ mod $4$

and by the reduction to the compact case, and the anisotropic kernel of $\mathfrak{so}_{p,q}$ being $\mathfrak{so}_{q-p}$, we get

both half-spin rep. of $\mathfrak{so}_{p,q}$ real $\Leftrightarrow$ $q-p \equiv 0$ mod $8$
both half-spin rep. of $\mathfrak{so}_{p,q}$ quaternionic $\Leftrightarrow$ $q-p \equiv 4$ mod $8$

and that again matches the tables in the Wikipedia article.
