How can one construct a strict quasigroup with neither a left nor a right identity on $n$ elements? When I say $Q$ is a strict quasigroup, I only mean that $Q$ is a quasigroup that is neither a loop nor is it associative. 
In general, I can find a quasigroup on $n$ elements with neither a left nor a right identity. If I need an example of one of these I generally just look at the Cayley table for the cyclic group of order $n$:
$$\begin{array}{c|cccc}
 & 1 & 2 & \cdots & n\\ \hline
      1 & 1 & 2 & \cdots & n\\
      2 & 2 & 3 & \cdots & 1\\
      \vdots & \vdots & \vdots & \ddots & \vdots\\
      n & n & 1 & \cdots & n-1\\
\end{array}
$$
Swap the first $2$ rows:
$$\begin{array}{c|cccc}
 & 1 & 2 & \cdots & n\\ \hline
      1 & 2 & 3 & \cdots & 1\\
      2 & 1 & 2 & \cdots & n\\
      \vdots & \vdots & \vdots & \ddots & \vdots\\
      n & n & 1 & \cdots & n-1\\
\end{array}
$$
Then swap the first $2$ columns:
     $$\begin{array}{c|cccc}
& 1 & 2 & \cdots & n\\ \hline
  1 & 3 & 2 & \cdots & 1\\
  2 & 2 & 1 & \cdots & n\\
  \vdots & \vdots & \vdots & \ddots & \vdots\\
  n & 1 & n & \cdots & n-1\\
\end{array}
$$ 
And the result is a quasigroup with neither a left nor a right identity. The problem is that the quasigroup that results from this process is in general associative, and there is no way to tell if an operation is not associative simply by looking at a Cayley table. So I am wondering, if I want to construct a strict quasigroup on $n$ elements which has neither a left nor a right identity, how would I go about doing that? 
It feels like this would be something simple to do, but so far I haven't been able to!
Any advice is greatly appreciated!
 A: Instead of swapping the first two columns, swap the first and third column.  Then there will still be no identity on either side since $2\cdot 2=2$ so $2$ is the only possible identity but $2\cdot 1=3$ and $1\cdot 2=3$.  Moreover, the quasigroup is not associative because (for instance) $$(1\cdot 2)\cdot 2=3\cdot 2=4$$ (or $1$ if $n=3$) and $$1\cdot (2\cdot 2)=1\cdot 2=3.$$
Note that this works only if $n\geq 3$.  If $n\leq 1$ then obviously it is impossible and for $n=2$ it is easy to see that any quasigroup must have an identity element.
A: Another example of the same kind is from the Cayley table of the subtraction $\mod n$, for $n\ge3$, a quasigroup operation where $0$ is a right identity and there is no left identity element. For example, for $n=4$ we have:
$$
\begin{array}{c|cccc}
-_{4} & 0 & 1 & 2 & 3\\
\hline
0 & \color{red}{0} & 3 & 2 & 1\\
1 & \color{red}{1} & 0 & 3 & 2\\
2 & \color{red}{2} & 1 & 0 & 3\\
3 & \color{red}{3} & 2 & 1 & 0\\
\end{array}
$$
The right identity is highlighted in red. Now swapping, for example, the $(n-1)$-th line and the $n$-th (here the third and the fourth) we get:
$$
\begin{array}{c|cccc}
\bullet & 0 & 1 & 2 & 3\\
\hline
0 & 0 & 3 & 2 & 1\\
1 & 1 & 0 & 3 & 2\\
3 & 3 & 2 & 1 & 0\\
2 & 2 & 1 & 0 & 3\\
\end{array}
$$
