Model
$$ y = \begin{pmatrix} y_1\\ \vdots \\ y_n \end{pmatrix} \in \mathbb{R}^n,\ X = \begin{pmatrix} 1 & x_{11} & \cdots & x_{1p} \\ 1 & x_{21} & \cdots & x_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n1} & \cdots & x_{np} \end{pmatrix},\\ x_i=(x_{i1},\ldots,x_{ip})^T,\ \bar{x}_+ = (\bar{x}_1, \ldots, \bar{x}_p)^T,\\ \bar{x}_j = \frac{1}{n} \sum_{i=1}^n x_{ij} \ (j=1, \ldots, p),\ \bar{y}= \frac{1}{n}\sum_{i=1}^ny_i,\\ \beta = (\beta_0, \beta_1, \ldots , \beta_p)^T,\ \beta_+ = (\beta_1, \ldots, \beta_p)^T $$
Setup
Now we consider $\tilde{x}_{ij} = a_j x_{ij} + b_j,\ \tilde{y}_i = cy_i + d$ where $a_j, b_j, c, d \in \mathbb{R}, a_j, c \neq 0, j=1,\ldots, p$.
Notation:
$$ A = \mathrm{diag}(1, a_1, \dots, a_p),\ b = (0, b_1, \ldots, b_p)^T $$
Question
We want to express the least-squares estimator $\tilde{\beta}=(\tilde{\beta}_0, \tilde{\beta}_1, \ldots, \tilde{\beta}_p)^T$ of the regression analysis for the transformed data using the least-squares estimator $\hat{\beta}=(\hat{\beta}_0, \hat{\beta}_1, \ldots, \hat{\beta}_p)^T$ of the regression analysis for the data before the transformation.
What I know
$$ \hat{y} = X\hat{\beta},\ \hat{\beta} = (X^TX)^{-1}X^Ty \leftarrow\mathrm{OLS\ estimator},\\ \tilde{X} = XA + \mathbb{1}_n b^T,\\ ||y-X\beta||^2 = \sum_{i=1}^n \{(y_i-\bar{y})-\beta_+^T(x_i-\bar{x}_+)\}^2 +n(\bar{y}-\beta_+^T\bar{x}_+ -\beta_0)^2. $$
If $\tilde{X} = XA,\ \tilde{y} = cy$ (i.e. only scaling), $$ \tilde{\beta} = (AX^TXA)^{-1}A^TX^Tcy = cA^{-1}(X^TX)^{-1}X^Ty = cA^{-1}\hat{\beta} = c\times \left(\hat{\beta}_0, \frac{\hat{\beta}_1}{a_1}, \ldots, \frac{\hat{\beta}_p}{a_p} \right) $$
Now that I am also translating, I do not know how to deform it. Do you have any ideas?