I would like to see an explicit example of two smooth isometric embeddings (in the Riemannian sense) $i_1,i_2:\mathbb{R}^2 \to \mathbb{R}^3$ such that there is no isometry $\varphi:\mathbb{R}^3\to \mathbb{R}^3$ suct that $i_2=\varphi\circ i_1$.
(I take here $\mathbb{R}^2,\mathbb{R}^3$ to be endowed with the standard Euclidean metrics).
Also, does there exist a pair of embeddings $i_1,i_2$ such that their images are not equivalent, i.e there is no isometry $\varphi:\mathbb{R}^3\to \mathbb{R}^3$ suct that $\text{Image}(i_2)=\varphi (\text{Image}(i_1))$?
Take a unit speed curve $\alpha \colon \mathbb{R} \rightarrow \mathbb{R}^3$ whose image lies in the $xy$ plane and consider the map $\varphi \colon \mathbb{R}^2 \rightarrow \mathbb{R}^3$ given by
$$ \varphi(s,t) = \alpha(s) + te_3 $$
where $e_3 = (0,0,1)$. The image of $\varphi$ is a cylinder over $\alpha$ with axis $e_3$. The differential of $\varphi$ is $(\alpha'(s), e_3)$ and since $\alpha'(s),e_3$ are orthogonal to each other and of unit length, this is an isometric immersion. If $\alpha$ is an embedding, you'll get an embedding. By choosing $\alpha$ appropriately, it is clear you can get an embedding whose image is not congruent to a plane (which itself can be considered as a cylinder over a line with axis that lies in the plane and s perpendicular to that line).