One to one mapping from $A(S_1)$ into $A(S_2)$ 
Let $S_1$ and $S_2$ be two sets. Suppose there exists a one to one mapping $\phi$ of $S_1$ into $S_2$. Show that there exists an one to one mapping from $A(S_1)$ into $A(S_2)$, where $A(S)$ means the set of all one to one mappings of $S$ onto itself.

Suppose $\{\phi_1,\phi_2,....\}$ are the elements of $A(S_1)$ and $\{\phi'_1,\phi'_2,....\}$
are the elements of $A(S_2)$. I thought the way to connect them lies through $'\phi'$. 
Let $F: A(S_1) \to A(S_2)$ defined by $F(\phi_i)=\phi\circ \phi_i$.
Not sure whether $ \phi\circ \phi_i=\phi'_i$??
 A: You want to induce the mapping some how.  Define $\phi^* : A(S_1) \to A(S_2)$ as $g \mapsto $ something related to $\phi$.  Well we know that $\phi \circ g$ is one-to-one and maps $S_1 \to S_2$.  What we need is a map from $S_2$ to itself though.  The only apparent option seems to be $\phi^{\leftarrow}$, which is a well-defined map since if $\phi^{\leftarrow}(a) = \{b,c\} \cup \dots$, then since $\phi\circ \phi^{\leftarrow} (a) = a$, we have $b = c$, by 1-1ness of $\phi$.  Iow, use the fact that $\phi$ is 1-1 to produce the fact that its reverse map is a function (not multi-valued, aka $\phi^{-1}$ returns singletons when provided with a singleton, etc).  You could use the notation $\phi^{-1} \equiv \phi^{\leftarrow}$ if you want, just know that there is a unique, well-defined map for $\phi$ also known as a right inverse.  Then we're almost done.  Now we've connected $\phi$ to the problem by defining $\phi^{-1}$ and showing that it's unique and well-defined.  Now 
define $\phi^* : A(S_1) \to A(S_2)$ as $\phi^*(f) = \phi f\phi^{-1}$.
Now show that $\phi^{-1}$ additionally is 1-1 and you have a composition of 1-1 maps which is 1-1.  $\phi^{-1}(a)  = \phi^{-1}(b) \implies \phi\phi^{-1}a = \phi\phi^{-1}b = a = b$.  So now we're done.
A: Given $\phi:S_1\rightarrow S_2$ and  $\psi:S_1\rightarrow S_1$ you are looking for a function  a function $\eta :S_2\rightarrow S_2$.
For sure, your $F(\psi)=\phi\circ g$ does not make sense..
You need $F(\psi)$ to be a function from  $S_2$ to $S_2$ 
i.e., given $s\in S_2$ you want to see what $F(\psi)(s)$ would be?
as you defined, $F(\psi)(s)=(\phi\circ \psi)(s)=\phi(\psi(s))$ does not make sense because $\psi(s)$ is not defined for $s\in S_2$ as $\psi\in A(S_1)$.
At present you only have  :

a function whose image is in $S_2$ namely $\phi$ but not a function whose domain is $S_2$

So, only possible option is to consider inverse (I strongly feel there is no other way)
so, $\phi^{-1}:S_2\rightarrow S_1$
So, you took an element in $S_2$ and landed in $S_1$ but you want an element in $S_2$ 
So, only option is to search for a function which takes elements in $S_1$ and gives elements in $S_2$.
You do have $\phi:S_1\rightarrow S_2$ which takes elements in $S_1$ and produce an element in $S_2$ but that would not help you because you would then only get an identity function $\phi\circ \phi^{-1}$.
Another possible option is to consider $\phi\circ \psi$.. This takes elements in $S_1$ and gives elements in $S_2$
This should work I believe...
i.e., $F(\psi)=\phi\circ \psi\circ \phi^{-1}$
Rest is your part to check if it is one one...
