When does there exist a $g\in G$ such that $H = gH'g^{-1}$? Suppose $H$ and $H'$ are two finite subgroups of $G$. Do there exist theorems that give conditions so that $H$ and $H'$ are conjugate, that is, there exists $g\in G$ such that $H = gH'g^{-1}$?
Edit: Maybe this question is too general. What about if they are both finite groups of invertible matrices?
 A: I'll use $K$ instead of $H'$, and write $H^g$ for $g^{-1}Hg$.
One condition is that $H$ must be isomorphic to $K$.  The function $\theta_g:x\mapsto x^g$ is a homorphism which maps $H$ bijectively to $K$, so $H\cong K$.  (In fact, such functions are referred to as inner automorphisms of $G$, and form a group themselves.)
The number of groups $K$ such that $H$ can be related by an inner automorphism to $K$ is easy to calculate.  If we let $G$ act on its set of subgroups by $g\cdot H = H^g$, we see that the stabilizer of $H$ under this action is exactly those $g$ for which $H^g=H$ - that is, the normalizer $N_G(H)$.  The orbit of $H$, then, is all subgroups $K$ for which there exists a $g\in G$ so that $H^g=K$.  By the Orbit-Stabilizer theorem, there are then $[G:N_G(H)]$ such groups.
Notice that since these groups are all in an orbit together, it must be true that $H\sim K \Leftrightarrow \exists g\in G :H^g=K$ is an equivalence relation.  The orbits are referred to as subgroup conjugacy classes.  
Determining the subgroup conjugacy classes of groups of invertible matrices is, actually, a rather difficult problem, even in when all subgroups are finite (i.e. in $GL_n(\mathbb{F}_q)$, which you can read about here).  One standard result on finite subgroups of matrices in general is that any finite subgroup of $GL_n(F)$ must have a conjugate contained in $O_n(F)$, which is provable using invariant symmetric bilinear forms.
