It is known that the general linear group $GL_n(\mathbb Q_p)$ over the $p$-adic numbers has $GL_n(\mathbb Z_p)$ as a maximal compact subgroup and every other maximal compact subgroup of $GL_n(\mathbb Q_p)$ is conjugated to this one. (Unfortunately, I don't have any reference for this fact and so I don't know any proof.)
Q: Is this also true for $GL_n(\mathbb C_p)$ and $GL_n(\mathcal O)$, where $\mathcal O \subset \mathbb C_p$ is the integer ring?
As noted in a comment, $GL_n(\mathcal O)$ is not compact in this situation. Further:
There are no maximal compact subgroups in $GL_n(\mathbb C_p)$.
Proposition: Let $F$ be any field which is complete with respect to a nonarchimedean value $\lvert \cdot \rvert$ but not locally compact, and let $K$ be a compact subgroup of $GL_n(F)$. Then there is a larger compact subgroup $\tilde K \supset K$.
Proof: $F$ being not locally compact means that either its residue field $\mathcal{O}/m_{\mathcal O}$ is infinite, or its value group $\lvert F \rvert$ is dense in $[0, \infty)$ (or both). In either case one can see (compare constructions here) that the closed set $$ U:= \{x \in F: \lvert x-1 \rvert < 1\}$$ (the "principal units") contains sequences without convergent subsequences; hence, so does the closed set of scalar matrices $U_n := U \cdot Id_n \subset GL_n(F)$; meaning that there must be $x \in U_n \setminus K$. The closure of the subgroup generated by such $x$ is the subgroup $ X:= x^{\mathbb Z_p} \subset GL_n(F)$, which is compact and in the centre of $GL_n(F)$. In particular, $\tilde K := KX =$ the subgroup generated by $K$ and $X$ is compact.
Note: We used completeness of $F$ for the description and compactness of $X$ in the above proof. K. Conrad's Compact Subgroups of $GL_n(\overline{\mathbb Q_p})$, which gave me the idea for that step, implies that the result is also true for $F=$ the (non-complete!) algebraic closure of $\mathbb Q_p$, albeit for a different reason: Each compact subgroup of that is contained in $GL_n(E)$ for some finite extension $E \vert \mathbb Q_p$.