Result on D41: Indicator function
Indicator function under scaling of the argument

Let $\mathbb{R}^N$ be a D816: Euclidean real Cartesian product such that
 (i) $E \subseteq \mathbb{R}^N$ (ii) $I_E$ is an D41: Indicator function on $\mathbb{R}^N$ with respect to $E$ (iii) $x \in \mathbb{R}^N$
Let $a \in \mathbb{R}$ be a D993: Basic real number such that
 (i) $$a \neq 0$$
Then
 (1) $$I_E(a x) = I_{a^{-1} E} (x)$$ (2) $$I_E (a^{-1} x) = I_{a E} (x)$$

Let $\mathbb{R}^N$ be a D816: Euclidean real Cartesian product such that
 (i) $E \subseteq \mathbb{R}^N$ (ii) $I_E$ is an D41: Indicator function on $\mathbb{R}^N$ with respect to $E$ (iii) $x \in \mathbb{R}^N$
Let $a \in \mathbb{R}$ be a D993: Basic real number such that
 (i) $$a \neq 0$$
Then
 (1) $$I_E(a x) = I_{E / a} (x)$$ (2) $$I_E (x / a) = I_{a E} (x)$$

Let $\mathbb{R}^N$ be a D816: Euclidean real Cartesian product.
Let $I_E$ be an D41: Indicator function for $E \subseteq \mathbb{R}^N$ in $\mathbb{R}^N$.
Let $a, x \in \mathbb{R}$ each be a D993: Basic real number such that
 (i) $a \neq 0$
Then
 (1) $$I_E(a x) = I_{\frac{1}{a} E} (x)$$ (2) $$\textstyle I_E (\frac{1}{a} x) = I_{a E} (x)$$
Also known as
Indicator function undder scaling of the underlying set
Proofs

Let $\mathbb{R}^N$ be a D816: Euclidean real Cartesian product such that
 (i) $E \subseteq \mathbb{R}^N$ (ii) $I_E$ is an D41: Indicator function on $\mathbb{R}^N$ with respect to $E$ (iii) $x \in \mathbb{R}^N$
Let $a \in \mathbb{R}$ be a D993: Basic real number such that
 (i) $$a \neq 0$$
Using result R4587 and the definition of an indicator function, we have the chain of equivalencies $$\begin{split} I_E(a x) = 1 \quad & \iff \quad a x \in E \\ & \iff \quad x \in a^{-1} E \\ & \iff \quad I_{a^{-1} E}(x) = 1 \\ \end{split}$$ whence the first claim follows as a consequence of R2965: Indicator function is uniquely identified by its support. The second claim follows by substituting $a^{-1} \neq 0$ in the place of $a$. $\square$