Let $\Lambda\le \mathbb R^n$ be a lattice and $\lambda_k(\Lambda):=$ the $k$-th minimum of the lattice ($k=1,\dots,n$), namely the $k$-th shortest distance from a lattice $\Lambda$ to the origin (please let me know if this definition of $\lambda_k(\Lambda)$ coincides with your usual definition of “$\lambda_k(\Lambda)$”). Note $\lambda_{1}(\Lambda) \le \dots \le\lambda_n(\Lambda)$. Now let $g\in GL(n,\mathbb R)$.
I wonder what the lower and upper bounds of $\lambda_k(g\Lambda)$ are in terms of $\lambda_j(\Lambda)$'s and $g$. I $\textit{speculate}$ something like the following may hold:
$$L(g) \lambda_{1}(\Lambda) \le \lambda_k(g\Lambda)\le U(g)\lambda_n(\Lambda),\forall g\in GL(n,\mathbb R),~\text{lattice}~\Lambda,$$
but not sure how to find appropriate positive functions $L(g),U(g)$. Please let me know the correct form of the above inequality as well as how to verify it.
Updates on my thought:
(1) I realized that $\lambda_k(g\Lambda)$ should vary w.r.t. how "orthogonal" the matrix $g$ is. In particular, if $g$ is orthogonal then $\lambda_{1}(\Lambda) \le \lambda_k(g\Lambda)\le \lambda_n(\Lambda)$. So $\det(g)$ shouldn't be the only thing I need to look at.
(2) I believe $L(g)$ and U(g) should be related to the eigenvalues of $g$, which roughly determine how vectors are "stretched".
A matrix $g \in GL_n(\mathbb{R})$ has $n$ singular values $\sigma_1(g) \ge \sigma_2(g) \ge \dots \ge \sigma_n(g) > 0$ which have the property that
$$\sigma_1(g) = \sup_{ \| x \| = 1} \| gx \|, \sigma_n(g) = \inf_{\| x \| = 1} \| gx \|.$$
(Equivalently, $\sigma_1(g) = \| g \|$ and $\sigma_n(g) = \| g^{-1} \|^{-1}$.) They're the square roots of the eigenvalues of $g^T g$. This gives, very straightforwardly,
$$\sigma_n(g) \lambda_1(\Lambda) \le \lambda_k(g \Lambda) \le \sigma_1(g) \lambda_n(\Lambda)$$
and probably something smarter can be said also but this is all you asked for.
The singular values can differ quite drastically from the eigenvalues; consider, for example, the case that $g = I + N$ is unipotent, so all of its eigenvalues are equal to $1$ but its singular values can get arbitrarily large (we straightforwardly have $\sigma_1(g) \ge \sigma_1(N) - 1$).