In Euclidean space $\mathbb{R}^d$ there are ${d+1 \choose 2}$ independent isometries (translations and rotations). In other words the dimension of the Euclidean isometry group is ${d+1 \choose 2}$.
In the $p$-adic space $\mathbb{Q}_p^d$ what do isometries look like? How many are there? Does it depend on $p$ as well as $d$?
Even in the simplest case, the isometry group of $\Bbb Z_2$ is really really big.
For starters, the space of linear isometries has infinitely many degrees of freedom. And that's just linear isometries. Same with the analytic (power series) isometries. And both are a drop in the ocean of general (non-linear, non-analytic) isometries
In the euclidean case, the image of a regular tetraedron (or $d+1$ points in general), completely determines the isometry. Or put another way, every point in euclidean space is uniquely determined by its distance to the $d+1$ points of the tetraedron. When you intersect spheres, eventually it's enough to pinpoint a point exactly.
In $p$-adic spaces, those "spheres" are not so nice. For starters, a sphere $S(x,r) = \{y \mid d(x,y) = r\}$ is not an algebraic variety anymore. When $r>0$, instead of being a $d-1$-dimensional variety, it's a large chunk of the $p$-adic space. In fact it's an open subset of $\Bbb Z_p^d$ and so it has nonzero density.
Since nontrivial spheres are open, no matter how many spheres $S(x_k,r_k)$ you pick, if all the $r_k$ are nonzero, the set $\{y \mid d(x_1,y) = r_1 \land d(x_2,y) = r_2 \land \ldots d(x_n,y) = r_n\}$ is an open subset, so if it is non-empty it can't be a singleton.
Which means that given $n$ points in $p$-adic spaces, the only other points that are uniquely determined by their distance to those $n$ points, are only those $n$ points themselves. And so the image of $n$ points can never determine an isometry completely (very far from it)
For example, if $f$ is a linear isometry of $\Bbb Z_2$, $f$ is determined by $f(1),f(2),f(4),\ldots$ If you know $f(1),\ldots f(2^{n-1})$, then looking at $f(2^n)$, its first $n$ binary digits have to be $0$, and the $n+1$th one has to be $1$, but that's the only thing you know about it. So you have to choose the whole $2$-adic integer $(f(2^n)-2^n)2^{-n-1}$. And so on. That's infinitely many "dimensions" right there.