Q:

Soft question, matching multiplication tables

I'm having troubles with an excercise from "Algebra: Chapter 0"

Both $(\mathbb{Z}/5\mathbb{Z})^*$ and $\mathbb{Z}/12\mathbb{Z}$ * consist of 4 elements. Write their multiplication
  tables, and prove that no re-ordering of the elements will make them match.

$(\mathbb{Z}/n\mathbb{Z})^*$ is multiplicative group of integers modulo n.
$(\mathbb{Z}/5\mathbb{Z})^* = \{[1]_5,[2]_5,[3]_5,[4]_5\}$ and $(\mathbb{Z}/12\mathbb{Z})^* =
\{[1]_{12}, [5]_{12}, [7]_{12}, [11]_{12}\}$
Here are the tables
$\begin{array}{|c|c|c|c|c|}
\hline
 *   &[1]_5& [2]_5 & [3]_5 & [4]_5    \\ \hline
[1]_5&  [1]_5 & [2]_5 & [3]_5 & [4]_5 \\ \hline
[2]_5&  [2]_5 & [4]_5 & [1]_5 & [3]_5 \\ \hline
[3]_5&  [3]_5 & [1]_5 & [4]_5 & [2]_5 \\ \hline
[4]_5&  [4]_5 & [3]_5 & [2]_5 & [1]_5 \\ \hline
\end{array}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{|c|c|c|c|c|}
\hline
 *   &[1]_{12}& [5]_{12} & [7]_{12} & [11]_{12}    \\ \hline
[1]_{12}&  [1]_{12} & [5]_{12} & [7]_{12} & [11]_{12} \\ \hline
[5]_{12}&  [5]_{12} & [1]_{12} & [11]_{12} & [7]_{12} \\ \hline
[7]_{12}&  [7]_{12} & [11]_{12} & [1]_{12} & [5]_{12} \\ \hline
[11]_{12}&  [11]_{12} & [7]_{12} & [5]_{12} & [1]_{12} \\ \hline
\end{array}$
But I do not really understand the task? What does it mean for tables to "match"? How do they match in this case? 
The only thing that comes to mind is that there are only two multiplication tables for any group of order $4$, one has single element of order $2$ and the other $3$. These two tables are representatives of such two types. I cannot really think of something else, what exactly should I prove and how? Could you help me?

A:

The right one always has one element on the diagonal, the left one does not. This is invariant under relabelling of the elements.
(alternatively, every element in the right group (Kleinse Viergruppe) has order 2, while this is not the case for the cyclic left group.

