Let $S\subset \mathbb{R}^3$ be an embedded surface and $g_S$ the induced metric from $\mathbb{R}^3$ onto $S$. Since isometries preserve Gaussian curvature, $S$ homogeneous $\Rightarrow S$ has constant Gaussian curvature. Is the reverse implication true?
[here homogeneous means that $\forall x,y\in S$ $\exists$ an isometry $f:S\to S$ with $f(x)=y$]
I think it's not true. My guess for a counter-example is the pseudosphere:
which is defined by: \begin{align*} x&=\text{sech}(u)\cos(v)\\ y&=\text{sech}(u)\sin(v)\\ z&=u-\tanh(u) \end{align*}
for $u\in\mathbb{R}$ and $v\in[0,2\pi)$.
$\textbf{Obs.:}$ To make it a proper smooth manifold, I'll consider the upper half $PS_+$ defined by $u>0$.
$PS_+$ has constant curvature $-1$, but I think (just think, can't prove it) that its only isometries are the rotations around the $z$-axis, which don't connect points $x,y$ with different heights.
Is it true that the rotations are the only isometries? Is there a simpler counter-example?
The pseudosphere is indeed a counterexample, and rotations are the only global isometries. It is, however, locally homogeneous in the sense that for any $x, y \in PS_+$ there is a local isometry between neighborhoods of $x, y$.
We can view the pseudosphere as an open subset of the hyperbolic plane (which cannot be globally embedded in $\Bbb R^3$), and this suggests a simpler counterexample: Any nonempty, proper subset of $\Bbb R^2$.
For example, if we take the punctured plane $P := \Bbb R^2 - \{0\}$, then then for any $p$ the length of the only geodesic starting at $p$ that cannot be extended forward (namely, $||p||$) is an invariant of the point $p$, so any isometry can only map points $p$ to points the same distance from $0$. In this case the isometries are precisely the rotations and reflections through $0$. This argument can be readily adapted to prove your assertion that the pseudosphere is also a counterexample.