This question is a continuation of my previous question: Uniqueness of codimension one embedding into the standard lattice.
Let $L$ be a positive-definite lattice of rank $n$, and suppose $L$ embeds into the standard lattice $\Bbb Z^{n+1}$. Is it true that the embedding is determined unique up to the following sense?
First, if $i:L\to \Bbb Z^{n+1}$ is an embedding of lattices, and if $f:\Bbb Z^{n+1}\to \Bbb Z^{n+1}$ is an automorphism, then $f\circ i$ is also an embedding. We regard $i$ and $f\circ i$ equivalent.
In the case that $L$ is decomposable, for example if $L=L_1\oplus L_2$, $g:L_1\to L_1$ is an automorphism and $i:X\to \Bbb Z^{n+1}$ is an embedding then $L=L_1\oplus L_2\xrightarrow{g\oplus \textrm{id}}L_1\oplus L_2=L\xrightarrow{i}\Bbb Z^{n+1}$ is a new embedding. We regard these two are equivalent.
For example, the lattice $(\Bbb Z,\langle 5\rangle)$ embeds into $\Bbb Z^2$ by $1\mapsto 2e_1+e_2$, and this is unique up to automorphism of $\Bbb Z^2$. (If the codimension is $>1$ then of course we do not have uniqueness.)
In the previous question that I linked, it is shown in the answer that such embedding is not unique, but a counterexample given has non-square-free determinant. Is the above question true if $\det L$ is square-free (so that $i(L)\subset \Bbb Z^{n+1}$ is a primitive lattice)?