Given a regular surface $A$, if one were to take the parallel surface i.e.$B = A+r\boldsymbol{n}$, where $\boldsymbol{n}$ is the unit normal to the surface $A$ and then isometrically deform the offset surface $B$ to a surface $C$, then would $C$ be isometric to the surface $E$ obtained by isometrically deforming $A$ to another surface $D$ and then taking the offset of $D$?
In other words, do the operations of offsetting and isometrically deforming commute with each other?
By Minding's theorem, two surfaces of the same Gaussian curvature are (locally) isometric. So consider the unit sphere and a non-umbilic surface of constant Gaussian curvature $1$. For example, take the parametrized surface $$\mathbf x(u,v) = \big(f(u)\cos v, f(u)\sin v, g(u)\big),$$ with $$f(u) = A\cos u,\ A>0, A\ne 1, \quad\text{and}\quad g(u) = \int_0^u\sqrt{1-A^2\sin^2t}\, dt.$$ (If $A>1$, the domain has to be appropriately restricted.) The principal curvatures (see, e.g., p. 53 of my differential geometry text) are $$k_1 = \frac{A\cos u}{\sqrt{1-A^2\sin^2u}} \quad\text{and}\quad k_2 = \frac1{k_1}.$$ Thus, the Gaussian curvature is still $1$, but the mean curvature $H=\frac{k_1+k_2}2$ is quite different from $1$.
Now the Gaussian curvature of the outward parallel surface at distance $r$ from the unit sphere is $$\frac1{(1+r)^2}$$ and the Gaussian curvature of the non-spherical surface (see here) is $$\frac1{1+2rH + r^2},$$ which is different. (The formulas in that linked question are working with the inward-pointing normal.) Thus, the parallel surface of the non-spherical surface is not isometric to the parallel surface of the spherical surface.