Fixed points under action of conjugate group elements Lecture notes:
If $g$ and $h$ are conjugate then $fix(g)=fix(h)$. To see this, say $g=k^{-1}hk$
If $g.s=s \iff k^{-1}hk.s=s \iff h.(k.s)=(k.s)$
Me: I understand the last line- I just fail to see how it implies $fix(g)=fix(h)$
EDIT: As Ben S says I think there was a typo in the notes and they forgot modulus signs so- New Aim: show $|fix(g)|=|fix(h)|$
 A: It doesn't. Consider for example the group of symmetries of a regular $n$-gon. All reflections are conjugate to each other, but each one fixes different points of the $n$-gon.
EDIT:
The argument in the post does show that $|\operatorname{fix}(g)| = |\operatorname{fix}(h)|$. Indeed, let a group $G$ act on a set $X$. Then for any element $k \in G$ the map $f_k: X \to X$ with $f_k(x) = k \cdot x$ is a bijection. If $g=k^{-1}hk$ then by your argument we have that $f_k(\operatorname{fix}(g)) = \operatorname{fix}(h)$. So the restriction of $f_k$ on $\operatorname{fix}(g)$ gives a bijection between $\operatorname{fix}(g)$ and $\operatorname{fix}(h)$.
A: To elaborate on what Amatai Yuval said:
Your argument already shows this.
Define $f:fix(g)\to fix(h)$ by $f(s)=k.s$. This is well-defined since you have shown that if $s\in fix(g)$, then $ks\in fix(h)$. If you can show $f$ is a bijection, you are done.
To this end: suppose $f(s_1)=f(s_2)$ for some $s_1,s_2\in fix(g)$. Then $k.s_1=k.s_2\implies k^{-1}k.s_1=s_2\implies s_2=s_2$. So, $f$ is injective.
Now let $t\in fix(h)$. We would like to show that $\exists s\in fix(g)$ such that $k.s=t$. Let $s=k^{-1}.t$. Using a similar argument to that in your question, you can show that $s\in fix(g)$, and $f(s)=k.s=t$. So, $f$ is surjective.
Hence, $|fix(g)|=|fix(h)|$.
A: As posilon says, we don't necessarily have $\mathrm{fix}(g)=\mathrm{fix}(h).$ However, the argument in the post does show, for example, that$$|\mathrm{fix}(g)|=|\mathrm{fix}(h)|.$$
