Are the roots of unity the only algebraic subgroups of the multiplicative group? $\newcommand{\G}{\mathbb{G}}$
Let $k$ a field (or maybe more generally an arbitrary ring with connected spectrum), $\G_m$ the multiplicative group over $k$. Are the $\mu_n = \{x^n = 1\}$ the only $k$-subgroup schemes of $\G_m$? What is the best way to see this? I have difficulty mostly in the algebraically closed field case.
My incomplete thoughts:
Working first on $\overline{k}$ an algebraically closed field, a proper closed subgroup-scheme $H$ of $\G_m$ must be affine of dimension zero, hence its $\overline{k}$-points are a finite discrete subgroup of $\overline{k}^\times$ and must coincide with $\mu_n$ for some $n$ (WLOG relatively prime to $p$, the characteristic).
How do I know there aren't any choices for nonreduced structure to put on these sets of points other than the ones which define $\mu_{np^k}$?
Edit: I suppose by homogeneity that the nonreduced structure on each point must be identical to that on the identity, so we reduce to classifying subgroups of $\G_m$ supported on the origin. Perhaps this can be approached ring-theoretically?
How do you descend the case above to $k_s$, so that Galois descent gives the case of an arbitrary field? The proof above doesn't seem to translate so well, as we could have to worry about $H$ being supported on points which aren't $k_s$-rational. But $k_s \to \overline{k}$ is a universal homeomorphism, so maybe one can sidestep the issue somehow?
I don't know how one would formally patch this result to arbitrary rings or schemes, but it seems intuitively true since the classification is uniform and discrete (so it doesn't seem like a subgroup could vary in families except to be constant.)
 A: Case of fields
Let me begin by confirming your suspicion over a field.

Claim (field case): Let $k$ be a field, and $H\subsetneq \mathbf{G}_{m,k}$. Then, $H$ is isomorphic to $\mu_{n,k}$ for some $n\in\mathbf{Z}_{\geqslant 1}$.

As the later answer uses more heavily the theory of multiplicative group schemes, let me sketch a proof of this using more the theory of finite group schemes.
Proof: If $k$ has characteristic $0$ then this is trivial. Indeed, in this case $H$ is etale by [Pink, Theorem 13.2]. Then, by [Pink, Theorem 12.2] we know that it suffices to show that $H(\overline{k})$ agrees with $\mu_N(\overline{k})$ as $\mathrm{Gal}(\overline{k}/k)$-modules for some $N$. But, as $H(\overline{k})$ is a $\mathrm{Gal}(\overline{k}/k)$-submodule of $\overline{k}^\times$ this is clear.
So, we may assume that $\mathrm{char}(k)=p>0$. As we have the connected-etale sequence
$$1\to H^\circ\to H\to \pi_0(H)\to 1,$$
(see [Milne, Proposition 5.58]), where $H^\circ$ is connected and $\pi_0(H)$ is etale, it suffices to show that $H^\circ=\mu_{p^r,k}$ for some $r$ and that $\pi_0(H)=\mu_{N,k}$ for some $p\nmid N$. Indeed, in this case, the connected-etale sequence necessarily splits by a simple computation (e.g. see [Milne, Corollary 15.40]), and as $H$ is commutative the claim will follow.
So, let us first assume that $H$ is etale. Then, the argument works exactly as in the characteristic $0$ case noting that the finite $\mathrm{Gal}(k^\mathrm{sep}/k)$-submodules of $(k^\mathrm{sep})^\times$ are of the form $\mu_N(k^\mathrm{sep})$ for some $p\nmid N$.
So, let us assume that $H$ is connected. Note then that the relative Frobenius map $F_H$ on $H$ is nilpotent (e.g. see [Pink, Proposition 15.6]), say $F_H^\ell=0$. Since $H\subseteq \mathbf{G}_{m,k}$ we see that this implies that $H\subseteq \ker(F_{\mathbf{G}_{m,k}}^\ell)=\mu_{p^\ell,k}$. But, we then see that $H\subseteq \mu_{p^\ell,k}$ gives us (by Cartier duality) a surjection $\underline{\mathbf{Z}/p^\ell\mathbf{Z}}_k\cong \mu_{p^\ell,k}^\vee\twoheadrightarrow H^\vee$. It's easy to see then that this implies that $H^\vee \cong \underline{\mathbf{Z}/p^r\mathbf{Z}}_k$ for some $r$, and so $H\cong H^{\vee\vee}\cong \mu_{p^r,k}$. $\blacksquare$

Issue with claim over arbitrary base
Unfortunately, the claim as stated is false over arbitrary base. In fact, we can write down a very simple example.
Let $R$ be a DVR with uniformizer $\pi$ (e.g. $R=\mathbf{Z}_p$ and $\pi=p$, or $R=\mathbf{C}[\![t]\!]$ and $\pi=t$) and consider
$$H:=V(\pi x^r-\pi)\subseteq \mathbf{G}_{m,R}=\mathrm{Spec}(R[x^{\pm 1}]).$$
Then, it's easy to see that
$$H_{R[\pi^{-1}]}\cong \mu_{r,R[\pi^{-1}]},\qquad H_{R/\pi R}\cong \mathbf{G}_{m,R/\pi R}.$$
In particular, $H\subseteq \mathbf{G}_{m,R}$, but $H$ is not isomorphic to $\mu_{n,R}$ for some $n$.
This group $H$ may look artificial, but it is in fact not. For example, let us consider the action
$$\mathbf{G}_{m,R}\times\mathrm{GL}_{2,R}\to\mathrm{GL}_{2,R},\qquad \left(x,\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right)\mapsto \begin{pmatrix}x^r & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}x^r & 0\\ 0 & 1\end{pmatrix}^{-1}.$$
Then,
$$H=\mathrm{Cent}\left(\mathbf{G}_{m,R},\left(\begin{smallmatrix}1 & \pi\\ 0 & 1\end{smallmatrix}\right)\right).$$
Fix a base scheme $S$ and $H\subsetneq\mathbf{G}_{m,S}$. By the field case discussed above, for every point $s$ of $S$ the subgroup $H_s\subseteq \mathbf{G}_{m,k(s)}$ is $\mu_{n_s,k(s)}$ for some $n_s\in\mathbf{Z}_{\geqslant 0}$ (where we take the convention that $\mu_{0,k(s)}:=\mathbf{G}_{m,k(s)}$). The issue is that there's no guarantee, as we saw in the above example, that the function
$$n\colon |S|\to\mathbf{Z}_{\geqslant 0},\qquad s\mapsto n_s,$$
is constant.
Challenge question: For $R$ as above, can you describe which functions $n\colon |\mathrm{Spec}(R)|\to \mathbf{Z}_{\geqslant 0}$ are actually realized by some $H\subsetneq \mathbf{G}_{m,R}$? Can you describe matrix realizations of the groups you construct?

The case of general base
To remedy the issue that $n_s$ is constant, one should imagine that we need $H\to S$ to have 'continuously varying fibers'. If you've been in the game long enough, you know that is code for 'flat'. Fore sxample, our example above is not flat as the map $R[x^{\pm 1}]/(\pi x^r-\pi)$ has $\pi$-torsion.
To state the result correctly, let us fix a base scheme $S$ and recall that a group $S$-scheme $H\to S$ is fppf if it is flat and locally of finite presentation. For simplicity let us assume that $S$ is locally Noetherian.${}^{\color{red}{(1)}}$

Claim (general base): If $H\subsetneq \mathbf{G}_{m,S}$ is an fppf group $S$-scheme, then $H$ is isomorphic to $\mu_{n,S}$ for some $n$.

Proof: Let us observe that as $H$ is fppf over $S$ that [Conrad, Corollary B.3.3] implies that $H\to S$ is of multiplicative type as in Definition B.1.1 of loc. cit. But, if $\xi\colon \text{Spec}(\Omega)\to S$ is a geometric point, then by[SGA 3, Exposé X, Théorème 7.1] there is an anti-equivalence of categories between group $S$-schemes of multiplicative type and discrete $\pi_1^\text{SGA3}(S,\xi)$-modules${}^{\color{red}{(2)}}$ finitely generated as an abelian group given by
$$H\mapsto X(H,\xi):= \text{Hom}_\Omega(H_\xi, \mathbf{G}_{m,\xi}).$$
In particular, from our closed embedding of group $S$-schemes $H\hookrightarrow \mathbf{G}_{m,S}$ we get a closed embedding of group $\Omega$-schemes $H_\xi\hookrightarrow\mathbf{G}_{m,\xi}$ which induces a surjection $X(\mathbf{G}_{m,S},\xi)\to X(H,\xi)$ which is evidently $\pi_1^\text{SGA3}(S,\xi)$-equivariant. But, $X(\mathbf{G}_{m,S},\xi)$ is isomorphic to $\mathbf{Z}$ with the trivial $\pi_1^\text{SGA3}(S,\xi)$-action. Thus, the surjection $X(\mathbf{G}_{m,S},\xi)\to X(H,\xi)$, which can't be an isomorphism as we assumed that $H\to \mathbf{G}_{m,S}$ was not an isomorphism, identifies $X(H,\xi)$ as $\mathbf{Z}/n\mathbf{Z}$ with the trivial $\pi_1^\text{SGA3}(S,\xi)$-action for some $n$. But, it's easy to compute that this is precisely $X(\mu_{n,S},\xi)$, and so the claim follows. $\blacksquare$
$\color{red}{(1)}$: That said, it should almost certainly be ok for locally topologically Noetherian $S$, and probably in general by a `descent to the Noetherian case' type argument. For instance, if $S$ is quasi-compact and quasi-separated then one quickly reduces to the Noetherian case by combining Tag 01ZA and Tag 01ZM.
$\color{red}{(2)}$: The topological group $\pi_1^\text{SGA3}(S,\xi)$ is an enlargement of the etale fundamental group $\pi_1^\text{et}(S,\xi)$ first studied in [SGA3, Exposé X, §6]. This group has its category of discrete sets with continuous action classifies the category $\mathbf{ULoc}(S_\text{et})$ -- the category of 'disjoint unions of' sheaves of sets on $S_\text{et}$ locally constant for the etale topology -- this is contrast to $\pi_1^\text{et}(S,\xi)$ whose category of discrete sets with continuoussets classifies $\mathbf{ULCC}(S_\text{et})$ -- the category of 'disjoint unions of' sheaves of sets on $S_\text{et}$ which are locally constant for the etale topology with finite fibers. The two groups actually coincide when $S$ is normal. You can also understand this as the maximal pro-discrete quotient of the pro-etale fundamental group from [BS].
References:
[BS] Bhatt, B. and Scholze, P., 2013. The pro-'etale topology for schemes. arXiv preprint arXiv:1309.1198.
[Conrad]Conrad, B., 2014. Reductive group schemes. Autour des schémas en groupes, 1(93-444), p.23.
[Milne] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.
[Pink] https://people.math.ethz.ch/~pink/ftp/FGS/CompleteNotes.pdf
[SGA3] M. Demazure, A. Grothendieck, Sch´emas en groupes I, II, III, Lecture Notes in Math 151, 152, 153, Springer-Verlag, New York (1970).
