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).
Best Answer
Let's take it apart bit by bit:
No. The $k$-points of $\mu_n$ are exactly the $n^{th}$ roots of unity inside $k$: these are the solutions over $k$ of the equation $T^n-1$. $\overline{k}$-points can include more stuff: think about the example when $k=\Bbb Q$ and $n=4$. The $k$-points are $\{-1,1\}$ and the $\overline{k}$-points are $\{1,-1,i,-1\}$ (under suitable identifications).
The problem here is that in characteristic $p$, we have $T^{p^m}-1=(T-1)^{p^m}$. So there is only one point in this group scheme, and it's not reduced. So, in particular, it's not just the trivial group scheme (which is what you would expect if you only looked at the set of $p^m$th roots of unity).