In Lyu12, Lemma 3.6 is as follows.
Lemma 3.6 For any non-negative integer $\alpha$ such that $gcd(2\alpha+1, q)=1$, there is a polynomial time reduction from the $SIS_{q, n, m, d}$ decsion problem to the $SIS_{q, n, m, (2\alpha+1)d+\alpha}$ decision problem.
Proof. To prove the lemma, we will show a transformation that maps the $SIS_{q, n, m, d}$ distribution to the $SIS_{q, n, m, (2\alpha+1)d+\alpha}$ distribution, and maps the uniform distribution over $\mathbb{Z}^{n\times m}_{q}\times \mathbb{Z}^{n}_{q}$ to itself. Given $(A, t)$, create a random vector $r \stackrel{\$}{\leftarrow} \{-\alpha, \cdot\cdot\cdot, 0, \cdot\cdot\cdot, \alpha \}^{m}$ and output $(A, (2\alpha+1)t+Ar)$. First observe that because $2\alpha+1$ is relatively prime to $q$, our transformation maps the uniform distribution to itself. And if $(A, t)$ canme form the $SIS_{q, n, m, d}$ distribution, then $(2\alpha+1)t+Ar=A((2\alpha+1)S+r)$, and since $s$ was chosen from the uniformly at random from $\{-d, \cdot\cdot\cdot, 0, \cdot\cdot\cdot, d \}^{m}$, it's not hard to see that $(2\alpha+1)S+r)$ is uniformly random in $\{-(2\alpha+1)d-\alpha, \cdot\cdot\cdot, 0, \cdot\cdot\cdot, (2\alpha+1)d+\alpha \}^{m}$.
I don't know the role of $gcd(2\alpha+1, q)=1$ in the proof. Why not create an problem $SIS_{q, n, m, d+\alpha}$ and reduces $SIS_{q, n, m, d}$ decsion problem to the $SIS_{q, n, m, d+\alpha}$ decision problem?