SVP=SIVP in ring lattice (ideal lattice)
Consider the ring $R = \mathbb{Z}[x]/(f(x))$ for $f(x)$ an irreducible monic polynomial of degree $n$.
Fix a basis $\{1,x,\dots,x^{n-1}\}$, and define the following transformation:
$$L(x^i) = x^{i+1}$$
Extend this transformation to all of $R$ via linearity.
Cayley-Hamilton theorem [1] states that the characteristic polynomial $p(t) = \det(L - tI)$ is such that:
$$p(L) = 0$$
One can explicitly compute the square matrix corresponding to $L$ (it is the companion matrix of $f(x)$), and it is known that:


The characteristic polynomial of $L$
The minimal polynomial of $L$
$f(x)$


Are all identical.
It follows that $f(L) = 0$, and this is the smallest polynomial $f(x)$ for which this holds, so among other things $\{1,x,\dots,x^{n-1}\}$ are linearly independent over $R$.

Now let $v(x)\in R$ be any vector (including a candidate "shortest vector").
We have that $\{v(x), xv(x), \dots, x^{n-1}v(x)\}$ are linearly independent over $R$.
This is because if there were some linear relation between them:
$$\sum_i a_iv(x)x^i = 0\implies v(x)\sum_i a_ix^i = 0\implies \sum_i a_i x^i = 0$$
Contradicting the linear independence.
As $\| x^i v(x)\|$ are all the same, we've exhibited a basis of vectors of the same norm.
So finding one short vector is enough to find all $n$ short vectors. 



[1] Technically we need to make sure that Cayley-Hamiltonian holds over modules over general rings. Fortunately it is known to hold for commutative rings, so we're fine.