Yes, this is true.
More generally, if $A$ is Lebesgue-measurable and $T$ is a linear mapping, then $\mu(T(A)) = \det(T)\mu(A)$. Also, if $T$ is a translation, then $\mu(T(A)) = \mu(A)$. This implies your statement since any dilation is a composite of a linear mapping and a translation.
See Theorem 2.20 of Rudin's _Real and Complex Analysis, 3rd Ed._