The question can be answered when one assumes an orthogonal matrix of predictors. Then the right singular vector matrix $\mathbf V$ equals the identity matrix $\mathbf I$, and the derivation below holds only for this case.
Consider the lasso problem
$$
\min_{\mathbf \beta} \frac12 ||\mathbf X \mathbf \beta - \mathbf y||^2_2 + \lambda ||\beta||_1
$$
Use the singular value decomposition
\begin{align}
\mathbf X &= \mathbf U \, \mathbf D\, \mathbf V^T\\
&= \sum_i \mathbf u_i \, d_i \mathbf v^T_i
\end{align}
and also expand $\beta$ and $\mathbf y$ as
\begin{align}
\beta &= \sum_i \beta_i \mathbf v_i\,\\
\mathbf y &= \sum_i y_i \mathbf u_i
\end{align}
Insert the whole stuff into the lasso functional to obtain
\begin{align}
\min_{\beta_i} \left(\sum_i \frac12 |\beta_i d_i - y_i|^2 + \lambda |\beta_i|\right)
\end{align}
In the SVD basis, the Lasso minimization problem thus decomposes into separate problems for each component.
For a vanishing singular value, $d_i=0$, the solution is $\beta_i=0$, so let's consider only the case $d_i >0$. For this, one can rewrite the previous equation as
$$\min_{\beta_i} \frac12 \left|\beta_i - \frac{y_i}{d_i}\right|^2 + \frac{\lambda}{d_i^2} \left|\beta_i\right|$$
As stated in the other answer, and also derived here, the solution of this problem is
$$
\beta^{Lasso}_i \ = \ \begin{cases} 0 &,\ |d_i y_i| \le \lambda \\ \frac{y_i}{d_i} - \frac{\lambda}{d_i^2} & ,\ |d_i y_i| > \lambda \end{cases}
$$
Compare this to the corresponding solution of the ordinary least squares problem and ridge regression (with regularization parameter $\alpha$):
\begin{align}
\beta_i^{OLS} &= \frac{y_i}{d_i}\\
\beta_i^{ridge} &= \frac{y_i}{\tilde d_i}, \quad \text{where}\ \ \tilde d_i = d_i \cdot \frac{d_i^2+\alpha}{d_i^2}
\end{align}
As is well known, ridge regression can be obtained by scaling a given singular value $d_i$ with a factor that depends on $d_i$ and $\alpha$.
On the other hand, the Lasso solution is more complex and also depends on the target $y_i$. With a good amount of enforcement, one can write the adjusted singular values as
$$
\beta_i^{Lasso} = \frac{y_i}{\bar d_i}, \quad \text{where} \ \ \bar d_i = d_i \frac{1}{\left(1 - \frac{\lambda}{d_i y_i}\right)\theta\big(|d_i y_i|-\lambda\big)}
$$
where $\theta$ is the Heaviside step function, which is zero for $|d_i y_i|<\lambda$ and otherwise one.
EDIT: Please note that for a general predictor matrix, the above statement is wrong, for two reasons: first and informally, if it would be correct, it would be the standard approach for an easy Lasso solution. Second, the formal reason why this doesn't work is that the L1-norm is not invariant under orthogonal transformations. So when one does the expansion of $\beta$ in two orthonormal basis sets, $\beta = \sum_i \beta_i \mathbf e_i = \sum_i b_i \mathbf v_i$, one has $\sum_i |\beta_i| \neq \sum_i |b_i|$.
So, what it does is not to solve the Lasso problem, but rather the more exotic Lasso problem
$$
\min_{\mathbf \beta} \frac12 ||\mathbf X \mathbf \beta - \mathbf y||^2_2 + \lambda ||\mathbf V^T \beta||_1
$$
