= Solution
A <plane> has both <principal curvature>[principal curvatures] zero, hence $K=H=0$. A <circular cylinder> of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, hence
$$
K=0,
\qquad
|H|=\frac1{2R}>0.
$$
Because the absolute value of <mean curvature> is preserved even if one reverses the chosen orientation, no <Euclidean motion> can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv0$ does not characterize planar pieces.
There are analogous examples for both signs of constant <Gaussian curvature>. Start with the arc-length <surface of revolution>
$$
X_A^+(u,v)
=\bigl(A\cos u\cos v,A\cos u\sin v,Z_A(u)\bigr),
\qquad
(Z_A')^2=1-A^2\sin^2u.
$$
On a sufficiently small interval about $u=0$, this is a regular surface and the <curvatures of an arc-length surface of revolution> give
$$
K=-\frac{(A\cos u)''}{A\cos u}=1.
$$
At $u=0$ its <mean curvature> is
$$
H_A^+(0)=\frac12\left(A+\frac1A\right).
$$
For $A=1$ this is a piece of the <unit sphere> and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.
For curvature minus one, use
$$
X_A^-(u,v)
=\bigl(A\cosh u\cos v,A\cosh u\sin v,W_A(u)\bigr),
\qquad
(W_A')^2=1-A^2\sinh^2u,
$$
again on a sufficiently small interval. Then
$$
K=-\frac{(A\cosh u)''}{A\cosh u}=-1,
\qquad
H_A^-(0)=\frac12\left(-A+\frac1A\right).
$$
The choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The <constant Gaussian curvature surfaces of revolution> therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.
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