<s>[INST] Showing that the OLS estimator is scale equivariant? I don't have a formal definition of scale equivariance, but here's what Introduction to Statistical Learning says about this on p. 217:

The standard least squares coefficients... are scale equivariant: multiplying $X_j$ by a constant $c$ simply leads to a scaling of the least squares coefficient estimates by a factor of $1/c$.

For simplicity, let's assume the general linear model $\mathbf{y} = \mathbf{X}\boldsymbol\beta + \boldsymbol\epsilon$, where $\mathbf{y} \in \mathbb{R}^N$, $\mathbf{X}$ is a $N \times (p+1)$ matrix (where $p+1 < N$) with all entries in $\mathbb{R}$, $\boldsymbol\beta \in \mathbb{R}^{p+1}$, and $\boldsymbol\epsilon$ is a $N$-dimensional vector of real-valued random variables with $\mathbb{E}[\boldsymbol\epsilon] = \mathbf{0}_{N \times 1}$.
From OLS estimation, we know that if $\mathbf{X}$ has full (column) rank, 
$$\hat{\boldsymbol\beta}_{\mathbf{X}} = (\mathbf{X}^{T}\mathbf{X})^{-1}\mathbf{X}^{T}\mathbf{y}\text{.}$$
Suppose we multiplied a column of $\mathbf{X}$, say $\mathbf{x}_k$ for some $k \in \{1, 2, \dots, p+1\}$, by a constant $c \neq 0$. This would be equivalent to the matrix
\begin{equation}
\mathbf{X}\underbrace{\begin{bmatrix}
1 & \\
  & 1 \\
  &  & \ddots \\
  &   &      & 1 \\
  &   &      & & c\\
  &   &      &  & & 1 \\
  &   &      &  & & &\ddots \\
  &   &      &  & & & & 1
  \end{bmatrix}}_{\mathbf{S}} = 
  \begin{bmatrix} \mathbf{x}_1 & \mathbf{x}_2 & \cdots & c\mathbf{x}_{k} & \cdots & \mathbf{x}_{p+1}\end{bmatrix} \equiv \tilde{\mathbf{X}}
\end{equation}
where all other entries of the matrix $\mathbf{S}$ above are $0$, and $c$ is in the $k$th entry of the diagonal of $\mathbf{S}$. Then, $\tilde{\mathbf X}$ has full (column) rank as well, and the resulting OLS estimator using $\tilde{\mathbf X}$ as the new design matrix is 
$$\hat{\boldsymbol\beta}_{\tilde{\mathbf{X}}} = \left(\tilde{\mathbf{X}}^{T}\tilde{\mathbf{X}}\right)^{-1}\tilde{\mathbf{X}}^{T}\mathbf{y}\text{.}$$
After some work, one can show that
$$\tilde{\mathbf{X}}^{T}\tilde{\mathbf{X}} = \begin{bmatrix}
\mathbf{x}_1^{T}\mathbf{x}_1 & \mathbf{x}_1^{T}\mathbf{x}_2 & \cdots & c\mathbf{x}_1^{T}\mathbf{x}_k & \cdots & \mathbf{x}_1^{T}\mathbf{x}_{p+1} \\
\mathbf{x}_2^{T}\mathbf{x}_1 & \mathbf{x}_2^{T}\mathbf{x}_2 & \cdots & c\mathbf{x}_2^{T}\mathbf{x}_k & \cdots & \mathbf{x}_2^{T}\mathbf{x}_{p+1} \\
\vdots & \vdots & \ddots & \vdots & \cdots & \vdots \\
c\mathbf{x}_k^{T}\mathbf{x}_1 & c\mathbf{x}_k^{T}\mathbf{x}_2 & \cdots & c^2\mathbf{x}_k^{T}\mathbf{x}_k & \cdots & c\mathbf{x}_k^{T}\mathbf{x}_{p+1} \\
\vdots & \vdots & \vdots & \vdots & \cdots & \vdots \\
\mathbf{x}_{p+1}^{T}\mathbf{x}_1 & \mathbf{x}_{p+1}^{T}\mathbf{x}_2 & \cdots & c\mathbf{x}_{p+1}^{T}\mathbf{x}_{p+1} & \cdots & \mathbf{x}_{p+1}^{T}\mathbf{x}_{p+1} \\
\end{bmatrix}$$
and
$$\tilde{\mathbf{X}}^{T}\mathbf{y} = \begin{bmatrix}
\mathbf{x}_1^{T}\mathbf{y} \\
\mathbf{x}_2^{T}\mathbf{y} \\
\vdots \\
c\mathbf{x}_k^{T}\mathbf{y} \\
\vdots \\
\mathbf{x}_{p+1}^{T}\mathbf{y}
\end{bmatrix}$$
How do I go from here to show the claim quoted above (i.e., that $\hat{\boldsymbol\beta}_{\tilde{\mathbf{X}}} = \dfrac{1}{c}\hat{\boldsymbol\beta}_{\mathbf{X}}$)? It's not clear to me how to compute $(\tilde{\mathbf{X}}^{T}\tilde{\mathbf{X}})^{-1}$.
 [/INST] I figured this out after posting the question. If my work is correct, however, I misinterpreted the claim. The $\dfrac{1}{c}$ scaling only occurs on the one component of $\boldsymbol\beta$ corresponding to the column of $\mathbf{X}$ being multiplied by $c$.
Notice that $\mathbf{S}$, in the notation above, is a diagonal, symmetric $(p+1) \times (p+1)$ matrix and has inverse (because it is diagonal)
$$\mathbf{S}^{-1} = \begin{bmatrix}
1 & \\
  & 1 \\
  &  & \ddots \\
  &   &      & 1 \\
  &   &      & & \frac{1}{c}\\
  &   &      &  & & 1 \\
  &   &      &  & & &\ddots \\
  &   &      &  & & & & 1
  \end{bmatrix}\text{.}$$
Note that $(\tilde{\mathbf{X}}^{T}\tilde{\mathbf{X}})^{-1}$ is a $(p+1)\times(p+1)$ matrix. Let's suppose that 
$$(\mathbf{X}^{T}\mathbf{X})^{-1} = \begin{bmatrix}
\mathbf{z}_1 & \mathbf{z}_2 & \cdots & \mathbf{z}_k & \cdots & \mathbf{z}_{p+1}
\end{bmatrix}\text{.}$$
Then it follows that 
$$(\tilde{\mathbf{X}}^{T}\tilde{\mathbf{X}})^{-1} = [(\mathbf{X}\mathbf{S})^{T}\mathbf{X}\mathbf{S}]^{-1} = (\mathbf{S}^{T}\mathbf{X}^{T}\mathbf{X}\mathbf{S})^{-1} = (\mathbf{S}\mathbf{X}^{T}\mathbf{X}\mathbf{S})^{-1}=\mathbf{S}^{-1}(\mathbf{X}^{T}\mathbf{X})^{-1}\mathbf{S}^{-1}\text{.}$$
Hence,
$$\mathbf{S}^{-1}(\mathbf{X}^{T}\mathbf{X})^{-1} = \begin{bmatrix}
\mathbf{z}_1 \\
& \mathbf{z}_2 \\
& & \ddots \\
& &  & \frac{1}{c}\mathbf{z}_k \\
& &  & & \ddots \\
& & & & & \mathbf{z}_{p+1}
\end{bmatrix}$$
and multiplying this by $\mathbf{S}^{-1}$ has a similar effect to what multiplying $\mathbf{X}$ by $\mathbf{S}$ did - it remains the same, except $\frac{1}{c}\mathbf{z}_k$ is multiplied by $\frac{1}{c}$:
$$\mathbf{S}^{-1}(\mathbf{X}^{T}\mathbf{X})^{-1}\mathbf{S}^{-1} = \begin{bmatrix}
\mathbf{z}_1 \\
& \mathbf{z}_2 \\
& & \ddots \\
& &  & \frac{1}{c^2}\mathbf{z}_k \\
& &  & & \ddots \\
& & & & & \mathbf{z}_{p+1}
\end{bmatrix}\text{.}$$
Therefore,
$$\begin{align}
\hat{\boldsymbol\beta}_{\tilde{\mathbf{X}}}&=\mathbf{S}^{-1}(\mathbf{X}^{T}\mathbf{X})^{-1}\mathbf{S}^{-1}(\mathbf{X}\mathbf{S})^{T}\mathbf{y} \\
&= \begin{bmatrix}
\mathbf{z}_1 \\
& \mathbf{z}_2 \\
& & \ddots \\
& &  & \frac{1}{c^2}\mathbf{z}_k \\
& &  & & \ddots \\
& & & & & \mathbf{z}_{p+1}
\end{bmatrix}\begin{bmatrix}
\mathbf{x}_1^{T}\mathbf{y} \\
\mathbf{x}_2^{T}\mathbf{y} \\
\vdots \\
c\mathbf{x}_k^{T}\mathbf{y} \\
\vdots \\
\mathbf{x}_{p+1}^{T}\mathbf{y}
\end{bmatrix} \\
&= \begin{bmatrix}
\mathbf{z}_1\mathbf{x}_1^{T}\mathbf{y} \\
\mathbf{z}_2\mathbf{x}_2^{T}\mathbf{y} \\
\vdots \\
\frac{1}{c}\mathbf{z}_k\mathbf{x}_k^{T}\mathbf{y} \\
\vdots \\
\mathbf{z}_{p+1}\mathbf{x}_{p+1}^{T}\mathbf{y}
\end{bmatrix}
\end{align}$$
as desired.
</s>