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)	\begin{equation} I_E(a x) = I_{a^{-1} E} (x) \end{equation}
(2)	\begin{equation} I_E (a^{-1} x) = I_{a E} (x) \end{equation}

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)	\begin{equation} I_E(a x) = I_{E / a} (x) \end{equation}
(2)	\begin{equation} I_E (x / a) = I_{a E} (x) \end{equation}

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)	\begin{equation} I_E(a x) = I_{\frac{1}{a} E} (x) \end{equation}
(2)	\begin{equation} \textstyle I_E (\frac{1}{a} x) = I_{a E} (x) \end{equation}

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)	$a \neq 0$

Since one has the chain of equivalencies \begin{equation} \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} \end{equation} then the first claim follows as a consequence of R2965: Indicator function is uniquely identified by its support. The second claim follows by applying the first one with a constant $b : = 1 / a$. $\square$