Find two different latin squares of order $5$ In the following question I am trying to find two different Latin squares of order $5$
Latin Square #1
\begin{array} 
& &1 &2 &3 &4 &5 \\
&5 &1 &2 &3 &4  \\
&4 &5 &1 &2 &3 \\
&3 &4 &5 &1 &2 \\
&2 &3 &4 &5 &1 \\
\end{array}
This Latin square was made by filling in the first row by $1,2,3,4,5$ and then filling in the second row by shifting cyclically one position to the right, and so on...
Latin Square #2
\begin{array} 
& &1 &2 &3 &4 &5 \\
&2 &3 &4 &5 &1  \\
&3 &4 &5 &1 &2 \\
&4 &5 &1 &2 &3 \\
&5 &1 &2 &3 &4 \\
\end{array}
This Latin square was made by filling in the first row by $1,2,3,4,5$ and then filling in the second row by shifting cyclically one position to the left, and so on...
So my questions are
1) Are these 2 different Latin squares of order $5$
2) Are my explanations for forming each Latin square correct?

They are different, in the sense that they're unequal.
They are the same, in the sense that they are isotopic, meaning that there is a way to permute the rows, columns, and symbols of one to obtain the other.  In this case, we can permute the rows of the first one to obtain the second one.
The Cayley table of $\mathbb{Z}_5$ is the second example (with different symbols).  It has no $2 \times 2$ subsquares ($2 \times 2$ submatrices which are Latin squares) since, in a group table, that would imply a subgroup of order $2$, violating Lagrange's Theorem.  So we can get an inequivalent Latin square by creating one with a $2 \times 2$ subsquare, i.e., by completing
$$
\begin{bmatrix}
1 & 2 & \cdot & \cdot & \cdot \\
2 & 1 & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
to a Latin square.
(Your explanations for how your Latin squares were generated make sense to me.)
