(ii) if two regions are identical up to rotations, translations, and reflections, then their measures should be equal; and
(iii) the measure of a square (cube, hypercube, etc.) with side $1$ should be $1$.
We could begin to formalize this concept by attempting to measure all subsets of the set of real numbers $\mathbb{R}$, but it turns out that this is impossible, as I will now demonstrate.
Let us say that two real numbers are equivalent if their difference is rational. Construct a subset $N$ of the half-open interval $[0,1)$ in $\mathbb{R}$ such that no two distinct elements are equivalent (this is possible by the axiom of choice), and define
\begin{equation}
N_r = \{x+r : x \in N \cap [0,1-r)\} \cup \{x+r-1 : x \in N \cap [1-r,1)\},
\end{equation}
where $r\in\mathbb Q\cap[0,1)$.
It follows that all $x \in [0,1)$ belong to exactly one $N_r$, namely $r = x-y$ or $r = x-y+1$, where $y \in N$ such that $x$ is equivalent to $y$. Otherwise, if $x \in N_r \cap N_s$, there exist $y, z \in N$ such that $x$ is equivalent to both $y$ and $z$, implying by transitivity that $y$ is equivalent to $z$, which is a contradiction. Thus, the $N_r$ are disjoint.
Then, by (i) and (ii), the measures of $N$ and $N_r$ are equal for all $r$. Moreover, by (iii) and (i),
\begin{equation}
1=\mu\left(\left[0,1\right)\right)=\mu\left(\bigcup_rN_r\right)=\sum_r\mu\left(N_r\right)=\sum_r\mu\left(N\right),
\end{equation}
where $\mu(N)$ is the measure of $N$. However, the rightmost sum is either $0$ or $\infty$, which is a contradiction.
This problem is a consequence of our ambition to assign a measure to all of $2^{\mathbb R}$. Nevertheless, we can circumvent it by focusing on a particular subset of $2^{\mathbb R}$ rather than the whole, and this is much of the subject of measure theory.