In physics, homogenous is used to mean that some quantity does not change with position.
This is less general than the use of homogeneous in mathematics, I think.
Question: So how does one define a metric space which is "homogeneous" in the sense of physics, i.e. its (metric) geometry is the same at every point?
(E.g. hyperbolic, Euclidean, and spherical geometries.)
There are definitions of a homogeneous metric space (see here or here), but both seem to be more general than what I am trying to describe above.
Attempt: Given a metric space $(X,d)$, denote the isometry group by $Iso(X)$, and for each $x \in X$, denote the subgroup of $Iso(X)$ consisting of elements fixing $x$ ($f(x) = x, f \in Iso(X)$), by $Iso(X,x)$.
Then a metric space "has the same geometry at each point" if and only if, for any two points $x_1, x_2 \in X$, one has that $Iso(X,x_1) \cong Iso(X, x_2)$.
Euclidean space satisfies this condition, since $Iso(X,x) \cong O(n)$ for any point $x$ in $n$-dimensional Euclidean space (I think). I don't know if hyperbolic and spherical geometries satisfy this too.
The definition of a homogeneous metric space is a little stronger than what you say: a metric space $(X,d)$ is homogeneous if and only if for every $x_1,x_2 \in X$ there exists $f \in Iso(X)$ such that $f(x_1)=x_2$. This implies the identity that you asked for, namely that $Iso(X,x_1) \simeq Iso(X,x_2)$, because one obtains an isomorphism $$A : Iso(X,x_1) \to Iso(X,x_2) $$ using the "conjugation" formula, also called the "adjoint" formula $$A(g) = f \circ g \circ f^{-1} \in Iso(X,x_2) \quad\text{for each}\quad g \in Iso(X,x_1) $$
All the geometries that you mention --- spherical, Euclidean, and hyperbolic --- are homogenous in this stronger sense: one simply works in Euclidean geometry, in spherical geometry, or in some particular model of hyperbolic geometry, to directly construct a desired isometry taking any given point $x_1$ to any other given point $x_2$. Those geometries therefore all satisfy the identity that you ask for, namely $Iso(X,x) \simeq O(n)$ for each $x \in X$, because you can easily use the symmetry of the metric to verify this identity for one particular value of $x$: use the origin in Euclidean space; the north pole in spherical space with the metric expressed in spherical coordinates; or the center of the Poincare disc model of hyperbolic space.