I actually already did something similar in Exercise 2. Consider the structure $(\{a,b\};P)$ where $P$ varies. There are 16 possibilities, 10 being essentially different (there are no automorphisms between any two of them).

	$P$
$\mathfrak{A}_{0}$	$\emptyset$
$\mathfrak{A}_{1}$	$\{<a,a>\}$
$\mathfrak{A}_{2}$	$\{<a,b>\}$
$\mathfrak{A}_{3}$	$\{<a,a>,<a,b>\}$
$\mathfrak{A}_{4}$	$\{<a,a>,<b,a>\}$
$\mathfrak{A}_{5}$	$\{<a,a>,<b,b>\}$
$\mathfrak{A}_{6}$	$\{<a,b>,<b,a>\}$
$\mathfrak{A}_{7}$	$\{<a,a>,<a,b>,<b,a>\}$
$\mathfrak{A}_{8}$	$\{<a,a>,<a,b>,<b,b>\}$
$\mathfrak{A}_{9}$	$\{<a,a>,<a,b>,<b,a>,<b,b>\}$

Given any structure $\mathfrak{A}=(\{c,d\};P)$ of size 2, we determine $k$ , the number of pairs in $P$ . If $k=0$ , any bijective function $h:|\mathfrak{A}|\rightarrow|\mathfrak{A}_{0}|$ will work as an isomorphism. Similarly for $k=4$ and $\mathfrak{A}_{9}$ . If $k=1$ , then either we have $<c,c>$ or $<d,d>$ in $P$ , or $<c,d>$ or $<d,c>$ in $P$ . Correspondingly, there is an isomorphism of $\mathfrak{A}$ onto $\mathfrak{A}_{1}$ or $\mathfrak{A}_{2}$ . If $k=2$ , then let $m$ be the number of pairs in $P$ of the form $<x,x>$ . If $m=0$ , then $P=\{<c,d>,<d,c>\}$ , and we are in case $\mathfrak{A}_{6}$ . If $m=1$ , then we have one such pair, say, $<c,c>$ , and one more either $<c,d>$ or $<d,c>$ , correspondingly we are in case $\mathfrak{A}_{3}$ or $\mathfrak{A}_{4}$ . If $m=2$ , then $P=\{<c,c>,<d,d>\}$ , and we are in case $\mathfrak{A}_{5}$ . Finally, if $k=3$ , then there is one pair missing in $P$ , and if $m=1$ , we are in case $\mathfrak{A}_{7}$ , otherwise, $m=2$ , and we are in case $\mathfrak{A}_{8}$ . Here “we are in case $\mathfrak{A}_{i}$ ” means that there is an obvious bijective correspondence $h$ between $|\mathfrak{A}|$ and $\{a,b\}$ such that $h$ preserves $P$ .