Is this a kind of "sketching"?
Yes, you can match the eigenvalues.  It is easy to construct a matrix with any desired eigenvalues: for instance, create a diagonal matrix and put the desired values on the diagonal.  Thus, you can get an exact match, not just an approximate match.  Nothing fancy is needed.

No, matching the eigenvalues doesn't imply that the eigenvectors of $B$ are somehow related in any way to the eigenvectors of $A$.

Given a value $n$ you can easily construct a matrix $B$ whose spectral norm is $n$, so that too can be exactly matched.

You don't need the Johnson–Lindenstrauss lemma for these results.