Given an undirected and unweighted graph $G = (V, E)$, is it possible to find a mapping $f: V \rightarrow \mathbb{R}^k$ for some $k$ such that for every $i, j \in V$, $\|f(i) - f(j)\|_2^2 = \Delta(i, j)$, where $\Delta(i, j)$ is the shortest path length between $i$ and $j$ in $G$?
I have been testing a few counterexamples for which these isometric embeddings in $\ell_2$ fail to exist (for example, the 4-cycle), but in this case they work.
This is not possible in general.
The 4-cycle is actually helpful to consider: embedding it in $\mathbb{R}^k$ in the way you describe requires the images of all four vertices to be coplanar, forming a square (since the distances between adjacent vertices must be $1$ and those between the non-adjacent pairs must be $\sqrt{2}$).
Now consider the complete bipartite graph $K_{2,3}$. Let us call its vertices $a_1,a_2,b_1,b_2,b_3$. In any embedding $f$ satisfying your condition, the 4-cycles $(a_1,b_1,a_2,b_2)$, $(a_1,b_1,a_2,b_3)$ and $(a_1,b_2,a_2,b_3)$ would all have to be mapped to squares as above, but the first two of these conditions imply $f(b_2) = f(b_3)$, contradiction.
