Outer automorphism of $S_6$ and conjugate stabilizers Let $f:S_6 \mapsto S_6$ be an outer automorphism of $S_6$ and consider the subgroups $$G =  \{\pi \in S_6 \mid \pi(1) = 1\}$$ and $$H = \{\pi \in S_6 \mid f(\pi)(1) = 1\}.$$
I would like to show that $G$ and $H$ are not conjugate subgroups of $S_6.$ It feels like there should be a direct reason why this is so but I do not see it.
Working directly we can infer that if $G = xHx^{-1}$ then $$g \in G \iff xgx^{-1} \in H \iff f(xgx^{-1})(1) =1 \iff y f(g) y^{-1} (1) = 1$$ where in the last step I used the substitution $y= f(x).$
The last equivalence is almost what I need but I don't know how to deduce that this would imply that $f$ is not an outer automorphism.
 A: Since the outer automorphism group of $S_5$ is trivial, and $x$ conjugates $G$ to $H$, we can choose $x$ such that $xgx^{-1} = f(g)$ for all $g \in G$.
So, $g := f^{-1}c_x$ (where $c_x$ is the automorphism of $S_6$ induced by conjugation by $x$) is an automorphism of $S_6$ that fixes every element of $G$. 
So $g((1,2))$ is an element of order $2$ that commutes with every element of the stabilizer of $1$ and $2$ in $S_6$, and the only such element is $(1,2)$. Hence $f^{-1}c_x$ is the identity, contradicting the assumption that $f$ is an outer automorphism.
A: If $f$ is an automorphism which is not inner we "know" that it will send transposition to $3$-transpositions (because if $f$ is not inner it changes the profil of permutations and we know how...). Now suppose you have $H=xGx^{-1}$.
From a group action point of vue this means that :
$$H=xGx^{-1}=xStab_{S_6}(1)x^{-1}=Stab_{S_6}(x(1))=\{\pi\in S_6|\pi(x(1))=x(1)\} $$ 
Now, take a transposition $\tau$ which fixes $x(1)$ (this can always be done) then $\tau\in H$, on the other hand this means, by definition of $H$, that $f(\tau)(1)=1$. The last one is impossible because a $3$-transpositions cannot fix any element.
