How to prove that: If two binary operations are anti-isomorphic and one of them is associative then the second one also will be associative? We know what is called an anti-isomorphic operation on a set S.
it is just a one two one $ g $ function mapping from $S$ to $S$.
$ g: S \rightarrow S$.      and it satisfy this condition
$ g(xy)= g(y)g(x) $.   
let`s say $\circ$ and $*$ is an anti-isomorphic binary operation on a set S.
if $S=\{1,2,3\}$  or if cardinality of $S=3$, and if it is a finite set, 
and
$
\begin{array}{c|ccccc}
\circ &   &   &        \\ \hline
      &  1&  1&  1&   \\ 
      &  1&  2&  3&   \\
      &  1&  1&  1&
\end{array}
$
Then it is an associative operation and it has only one anti-isomorphic operation!
$
\begin{array}{c|ccccc}
* &   &   &        \\ \hline
      &  1&  1&  1&   \\ 
      &  1&  2&  1&   \\
      &  1&  3&  1&
\end{array}
$
Which is again associative!
We know that basically an anti-isomorphic operations Cayley tables are transpose matrices to each other.   
I need to know if a binary operation has a lot of anti-isomorphism operation, then will all of them associative?
Also what about if S is an infinite set or uncountable set?
If they will be associative as well then I need to know How to prove it? 

Let $g\colon (S,\circ)\to (S,*)$ be the antiisomorphism and $h\colon S\to S$ be its inverse. By assumption
$$
g(x\circ y)=g(y)*g(x)
$$
and it's easy to prove that $h$ is an antiisomorphism as well:
$$
h(x*y)=h(y)\circ h(x)
$$
(prove it).
Now, suppose $*$ is associative. Then
\begin{align}
x\circ (y\circ z)&=h(g(x))\circ(h(g(y)\circ h(g(z)))\\
&=h(g(x))\circ(h(g(z)*g(y)))\\
&=h(g(x))\circ(h(g(y\circ z)))\\
&=h(g(y\circ z)*g(x)))\\
&=h((g(z)*g(y))*g(x))\\
&=h(g(z)*(g(y)*g(x))) &&\text{by associativity of $*$}\\
&=\dots
\end{align}
