Finding the isomorphism from Inner automorphism group of $G$ to $\frac{G}{Z}$ Let $G$ be a finite group and $Z$ its center. If $I$ is the group of inner automorphisms of $G$ then $\frac{G}{Z}$ is isomorphic to $I$.
Now given $I$ is it possible to find a $g$ $\epsilon$ $\frac{G}{Z}$ such that it effects the automorphism produced by $T$ $\epsilon$ $I$.
My first approach was to look at  $H=\{x\epsilon\frac{G}{Z}:T(x)=x\}$. This forms a sub group of $\frac{G}{Z}$ and the $g$ I need is clearly in this group. Now this $g$ should also commute with all elements in this group. So I can look at the center of $H$.
I can't seem to get any further than this. I would appreciate any help if there is in general a possible way of finding such an element.
Further,
In case of the group of invertable matrices I believe the map $T(x)$ is equivalent to a similarity transformation and this is essentially finding the matrix R such that $RAR^{-1}=T(A)=B$ which is equivalent to the eigen value problem. So it shouldn't be impossible. But they are of course infinite groups. If possible I want to solve it without representations.
 A: Consider $x\in G$, and an inner automorphism $\alpha$ of $G$. We seek to find a the  $g\in G/Z(G)$, such that $gxg^{-1}=\alpha (x)$. Note that $g$ can be any representative from the relevant coset of $G$ with respect to the normal subgroup $Z(G)$, so in practice, it is probably wise to fix a $g$ for each non-identity coset. You can then check the actions of each of those representatives on $G$ via conjugation and thus construct the desired correspondence.
A: To proof your assertion, you should map $\phi:G\to{\rm Inn}G$ via 
$$\phi(g)=I_g$$ 
where $I_g:G\to G$ is defined as $I_g(x)=gxg^{-1}$, that is, $I_g$ is a inner autohomomorphism.
It isn't to hard to show that this $\phi$ is an epimorphism.
Now by the 1st homomorphism fundamental theorem 
$${\rm Inn}G\cong G/\ker\phi.$$
But it happens that $\ker\phi=Z$. Done.
A: That's pretty simple and there's not much to say: by definition $T$ is conjugation by some $g\in G$: $T=\phi(g)$, and the inverse isomorphism is given by $\;T\longmapsto \phi^{-1}(T)=gZ$.
