Real forms of $\mathbb{G}_m$ as a variety Notation: Denote $k=\mathbb{R}$, $K=\mathbb{C}$ and let $G:=\text{Gal}(K/k)=\{\text{id},\sigma\}$, where $\sigma$ is complex conjugation.
Background: In the Notices article A Gentle Introduction to Arithmetic Toric Varieties the author introduces (in slightly different notation) three varieties over $k$:

*

*$\mathbb{G}_m=\{uv-1=0\}$,

*$\text{S}^1=\{x^2+y^2-1=0\}$,

*$Y=\{a^2+b^2+1=0\}$.

He establishes that they are pairwise non-isomorphic: e.g. looking at the real points in Euclidean topology $\mathbb{G}_m(k)$ has two connected components, while $\text{S}^1(k)$ has only a single connected component and $Y(k)=\emptyset$ has no $k$-points at all.
He then explains that $\mathbb{G}_m$ and $\text{S}^1$ are twisted forms of each other, since they become isomorphic (as affine $K$-varieties) after base change to $K$, explicitly:
$$\varphi:\text{S}^1\times_kK\rightarrow\mathbb{G}_m\times_kK,\quad (x,y)\mapsto(u,v)=(x+iy,x-iy).$$
Explicitly the inverse morphism is given by:
$$\varphi^{-1}:\mathbb{G}_m\times_kK\rightarrow\text{S}^1\times_kK,\quad (u,v)\mapsto(x,y)=\left(\frac{u+v}{2},\frac{u-v}{2i}\right).$$
As far as I see $Y$ also becomes isomorphic to $\mathbb{G}_m$ and $\text{S}^1$ after base change to $K$, an isomorphism is given e.g. by:
$$\psi:\text{S}^1\times_kK\rightarrow Y\times_kK,\quad (x,y)\mapsto(a,b)=(ix,iy),$$
$$\psi^{-1}:Y\times_kK\rightarrow\text{S}^1\times_kK,\quad (a,b)\mapsto(x,y)=(-ia,-ib).$$
This would indicate that the complex affine variety $\mathbb{G}_m\times_kK$ has (at least) 3 distinct real forms, namely $\mathbb{G}_m, \text{S}^1$ and $Y$. Note that I do not consider the structure of a group variety, with which you could endow $\mathbb{G}_m, \text{S}^1$ in the usual way but not $Y$ as it is pointless (over $k$).
As is well known for classes of varieties $X$ that form a ‘stack’ (e.g. quasi-projective varieties - so affine $X$ is OK)
$k$-forms of some $X\times_kK$ are in bijection to $$\text{Tw}(X)=\text{H}^1_{\text{gal}}(k,\text{Aut}_{\text{var}}(X\times_kK))=\text{H}^1_{\text{grp}}(\text{Gal}(K/k),\text{Aut}_{\text{var}}(X\times_kK))$$ (cf., e.g., Gille/Szamuely, "Central Simple Algebras and Galois Cohomology", Thm. 2.3.3 and sec. 5.2 for special versions). The author claims
$$\text{Tw}(\mathbb{G}_m)=\{\mathbb{G}_m, \text{S}^1\}$$ only contains 2 distinct elements.
And indeed using this post (and using that morphisms of affine varieties are in bijection to algebra morphisms) we see that $\text{End}_{\text{var}}(\mathbb{G}_m\times_kK)\cong\left\{a\cdot t^n:a\in K^{\times},n\in\mathbb{Z}\right\}$ and therefore $A:=\text{Aut}_{\text{var}}(\mathbb{G}_m\times_kK)\cong\left\{a\cdot t^n:a\in K^{\times},n\in\mathbb{Z}^{\times}=\{\pm 1\}\right\}\cong K^{\times}\times\mathbb{Z}/2\mathbb{Z}$. The Galois action of $\text{G}$ on $A$ is induced by $\tilde{\sigma}:A\rightarrow A, (a,\pm 1)\mapsto(\sigma(a),\pm 1)=(\bar{a},\pm 1)$ using the identification with $K^{\times}\times\mathbb{Z}/2\mathbb{Z}$ just given.
Using the standard method to compute group cohomology for cyclic groups through norm and difference maps, $N=\sum\limits_{i=0}^{2-1}\tilde{\sigma}^i=\tilde{\sigma}+\tilde{\text{id}}$ and $\Delta=\tilde{\sigma}-\tilde{\text{id}}$ (cf., e.g., Gille/Szamuely, Ex. 3.2.9), as $$\text{H}^1_{\text{grp}}(\text{G},A)\cong\text{ker}(N)/\text{im}(\Delta)$$ this can easily seen to indeed be $\cong\mathbb{Z}/2\mathbb{Z}$.
Question: Since $3>2$ there must be a mistake in these considerations: What is this mistake and what is $\text{Tw}(\mathbb{G}_m)$?
Remark 1: As indicated above one could also look at the additional structure of group varieties. For this variant we have, if I am not mistaken, $B:=\text{Aut}_{\text{grp-var}}(\mathbb{G}_m\times_kK)\cong\left\{1\cdot t^n:n\in\mathbb{Z}^{\times}=\{\pm 1\}\right\}\cong \mathbb{Z}/2\mathbb{Z}$ which yields $\text{Tw}_{\text{grp-var}}(\mathbb{G}_m)=\text{H}^1_{\text{grp}}(\text{G},B)\cong\mathbb{Z}/2\mathbb{Z}$. Since $Y$ has no $k$-point it cannot have the structure of a group variety, thus $Y$ is no real form of the complex group variety $\mathbb{G}_m\times_kK$ and there is no contradiction from the above arguments. The issue therefore lies in classifying real forms of $\mathbb{G}_m\times_kK$ as only affine $k$-varieties without any (potential) group structure.
Remark 2: I choose the notation $K/k$ to indicate that analogous issues should also arise for e.g. $\mathbb{Q}(i)/\mathbb{Q}$ the Gaussian rationals and probably even in positive characteristic (probably $\neq 2$, since $\varphi^{-1}$ requires inversion of $2$).
 A: As Alex Youcis encouraged me to do so, I try to answer myself.
The mistake was that $A:=\text{Aut}_{\text{var}}(\mathbb{G}_m\times_kK)\cong\left\{a\cdot t^n:a\in K^{\times},n\in\mathbb{Z}^{\times}=\{\pm 1\}\right\}\not\cong K^{\times}\times\mathbb{Z}/2\mathbb{Z}$ (an abelian group) but rather $\cong K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}$ (a non-abelian group). Explicitly the group composition is given by $(a_1,b_1)\cdot(a_2,b_2)=(a_1\cdot a_2^{b_1},b_1\cdot b_2)$.
Since the $N$-$\Delta$-approach only computes group cohomology when the $G$-set in question is an abelian group where the abelian group structure is compatible with the $G$-action (i.e. a $G$-module) we cannot compute $\text{H}^1_{\text{grp}}(\text{G},A)$ via $\text{ker}(N)/\text{im}(\Delta)\cong\mathbb{Z}/2\mathbb{Z}$.
The correct $\text{Tw}(\mathbb{G}_m)$ should be computed via non-abelian group cohomology as presented e.g. in J.S. Milne's lecture notes "Algebraic groups", sec. 27.a, pp. 469ff. We use 1-cocycles and equivalence $\sim$ of 1-cocycles as described there.
A 1-cocycle in this context is given by a pair of elements of $A\cong K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}$ of the form $c=(c_{\text{id}},c_{\sigma})$. The cocycle condition immediately yields $c_{\text{id}}=1_A=(1,1)$ and we will discard it from the notation and identify $c=c_{\sigma}=(a,b)=(s+ti,b)$. Exploiting the cocycle condition further we get the set of 1-cocycles to be $$\text{Cocyc}^1=\{(s+ti,1):(s,t)\in k^2\setminus\{(0,0)\} \wedge s^2+t^2-1=0\} \cup \{(s,-1):s\in k^{\times}\}$$ $$= \{(a,1):a\in\text{U
}(1)\}\quad\dot{\cup}\quad\{(a,-1):a\in\mathbb{R}^+\}\quad\dot{\cup}\quad\{(a,-1):a\in\mathbb{R}^-\}.$$
Straight forward computation yields that these three disjoint sets are exactly the 3 equivalence classes of 1-cocycles.
In summary $\text{H}^1_{\text{grp}}(\text{G},A)=\left(\text{Cocyc}^1/\sim\right)=\{\mathbb{G}_m,\text{S}^1,Y\}=\text{Tw}(\mathbb{G}_m)$
Alternatively Alex Youcis suggested to use the short exact sequence of non-abelian groups (recall that abelian groups are also non-abelian groups) $$1\rightarrow K^{\times} \rightarrow K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow 1,$$
which yields a long exact sequence of pointed sets, whose relevant part is given by
$$\ldots\rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}) \rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}) \rightarrow \text{H}^1_{\text{grp}}(\text{G},\mathbb{Z}/2\mathbb{Z}).$$
By the classic Hilbert's Theorem 90 (cohomological version) we have $\text{H}^1_{\text{grp}}(\text{G},K^{\times})\cong 1$ and by remark 1 from the question $\text{H}^1_{\text{grp}}(\text{G},\mathbb{Z}/2\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$, which makes the previous sequence more explicit
$$\ldots\rightarrow 1 \rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}) \xrightarrow{\alpha} \mathbb{Z}/2\mathbb{Z}.$$
$\text{H}^1_{\text{grp}}(\text{G},A)$ consists of the fibers above $\mathbb{Z}/2\mathbb{Z}\cong\{\mathbb{G}_m, \text{S}^1\}$ of which $\alpha^{-1}\left(\mathbb{G}_m\right)=\{\mathbb{G}_m\}$ must be a singleton, because of the properties of exact sequences of pointed sets. This leaves us with computing the other fiber $\alpha^{-1}\left(\text{S}^1\right)$.
Computing $\alpha^{-1}\left(\text{S}^1\right)$ explicitly seems to me not much simpler than computing $\text{H}^1_{\text{grp}}(\text{G},A)$ directly using 1-cocycles and equivalences as done above. At least I do not see how to do it differently, which gives this approach the feeling of a dead end. But more apt people might have better ideas than me.
Edit note: removed the trailing dots in the long exact sequences as suggested by Alex Youcis.
