connection between conjugation and "relabeling" If we look at $S_n$ we know that conjugation preserves cycle type, and that conjugation of some $\tau$ by $\sigma$ permutes the numbers in the cycle representation of $\tau$ through $\sigma$. The ''relabeling invariant'' property of a permutation is its cycle type, and it seems that this is exactly the conjugation invariant property of a permutation. With the group of linear bijections from some finite dimensional vector space to itself something similar seems to happen. Fixing a basis, any linear bijection induces a matrix, and if we look at the conjugacy class of that matrix we get exactly all other matrix representations of the original map when other choices of basis are made. Here , the "relabeling invariant" property seems to be the map that is represented, which is what conjugation (sort of) preserves, in the sense made precise above. Is the fact that conjugation has something to do with relabeling a general group theoretic phenomenon(i.e. is something similar true for any group)? Can one make the above notions precise? Do non-inner automorphisms have this property?
I am sorry if this question is somewhat vague or informal.

One common thing is that there are group actions (groups acting on some sets) in both cases: the action of of a group $G$ on itself by conjugation and the group of similarly transformations acting on the set of square matrices over some field.  In the former case, the orbits are the conjugacy classes.  In the latter case, the representatives of orbits are some matrices (such as diagonal matrices) which are called canonical forms.  
If $G$ is a permutation group in $S_n$, then $G$ acts on the $n$ points, and conjugating a permutation relabels its points (and preserves the cycle structure).  In the case of matrices, $X^{_-1}AX$ is a matrix representing the same linear transformation but with respect to a different basis.  But these matrices do not necessarily permute the basis elements or the points in a vector space (for example, the matrix $A$ can be singular).   
