Let $X$ be a positive definite lattice of rank $n$, and suppose $X$ embeds into the standard lattice $\Bbb Z^{n+1}$. Then is it true that the embedding is determined uniquely up to automorphism of $\Bbb Z^{n+1}$?
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.)
I think that I need an extra assumption that in the case that $X$ is decomposable. For example if $X=X_1\oplus X_2$ and $f:X_1\to X_1$ is an automorphism and $i:X\to \Bbb Z^{n+1}$ is an embedding then $X=X_1\oplus X_2\xrightarrow{f\oplus \textrm{id}}X_1\oplus X_2=X\xrightarrow{i}\Bbb Z^{n+1}$ is a new embedding. I think I should regard this kind of embeddings as the same as the original embedding.
The answer is NO.
The lattice $(\mathbb{Z},\langle 25\rangle)$ embeds in $\mathbb{Z}^2$ in two ways: define $f_i:\mathbb{Z}\to \mathbb{Z}^2$ by
$f_1(1)=5 e_1$ ,and $f_2(1)=3e_1+4e_2$, where $(e_1,e_2)$ is the canonical basis of $\mathbb{Z}^2$.
Then there is no automorphism $u$ of $\mathbb{Z}^2$ such that $f_2=u\circ f_1$.
Indeed if the matrix of $u$ with respect to $(e_1,e_2)$ is $U=\pmatrix{a&b\cr c& d}\in \mathrm{GL}_2(\mathbb{Z})$, then we would have $(u\circ f_1)(e_1)=5u(e_1)=5a e_1+5 ce_2=f_2(e_1)=3e_1+4e_2$. Since $a,c\in\mathbb{Z}$, we have a contradiction.