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**cubejunkies** Let* G *be a group and let *g* be some element of *G*. Given that $\displaystyle \tau$ is an isomorphism is such that $\displaystyle \tau(g) = g_1$, show that $\displaystyle S(g)$ and $\displaystyle S(g_1)$ are conjugate subgroups of *G*. Here, $\displaystyle S(x)$ is the subgroup of the set of automorphisms of *G* that fix the element *x*. In other words, $\displaystyle S(x) = \{\sigma\in Aut G \bigr| \ \sigma(x) = x\}$.

I haven't gotten anywhere with this after about 5 hours now. I feel like the key to the problem will be pulling in $\displaystyle \tau$ somehow- intuitively, *g* and *g*_{1} seem to me sort of like inverses of each other across the $\displaystyle \tau$ isomorphism. Since $\displaystyle \tau$ is a member of the group Aut(G), it has an inverse, say $\displaystyle \tau^{-1}$ such that $\displaystyle \tau^{-1}(g_1) = g$ and I feel like $\displaystyle \tau$ somehow will be very involved in the conjugation of $\displaystyle S(g)$. I don't know. I've been staring at a whiteboard basically writing the definition of conjugation over and over and over and over without getting anywhere. I already have proved that $\displaystyle S(g)=\{\sigma\in Aut G \bigr| \ \sigma(g) = g\}$ is a subgroup of Aut(G). I'm stuck :/