For elements $a,b \in G$, group that acts on the set $X$, show that $a$ and $b^{-1}ab$ have the same numbers of fixed points in $X$ The group $G$ acts on the finite set $X$.
Let $a,b \in G$.
Show that $a$ and $b^{-1}ab$ have the same number of fixed points in $X$

My idea was perhaps to define two sets which contain all the fixed points of $a$ and $b^{-1}ab$ respectively and then proceed to find a bijection between those groups. The function I thought of is defined by $f(x)=b^{-1}x$.
Would that be a sufficient proof and if so how exactly can I do that (for example is it necessary to prove it is well-defined)?

 A: \begin{alignat}{1}
\operatorname{Fix}(b^{-1}ab) &= \{x\in X\mid (b^{-1}ab)\cdot x=x\} \\
&= \{x\in X\mid b^{-1}\cdot((ab)\cdot x)=x\} \\
&= \{x\in X\mid (ab)\cdot x=b\cdot x\} \\
&= \{x\in X\mid a\cdot (b\cdot x)=b\cdot x\} \\
\end{alignat}
The map $f_b\colon X\to X$, defined by $x\mapsto b\cdot x$, is bijective. Therefore:
\begin{alignat}{1}
|\operatorname{Fix}(b^{-1}ab)| &= |\{x\in X\mid a\cdot (b\cdot x)=b\cdot x\}| \\
&= |\{b\cdot x\in X\mid a\cdot (b\cdot x)=b\cdot x\}| \\
&= |\{y\in X\mid a\cdot y=y\}| \\
&= |\operatorname{Fix}(a)| \\
\end{alignat}

Following your idea, we have the two sets $A:=\operatorname{Fix}(b^{-1}ab)=\{x\in X\mid a\cdot (b\cdot x)=b\cdot x\}$ (see above) and $B:=\operatorname{Fix}(a)=\{y\in X\mid a\cdot y=y\}$; define the map $f\colon A\to B$ by $x\mapsto y:=b\cdot x$.

*

*Good definition: $x\in A \Rightarrow a\cdot(b\cdot x)=b\cdot x \Rightarrow a\cdot y=y \Rightarrow y\in B$;

*Injectivity:

\begin{alignat}{1}
f(x)=f(x') &\Rightarrow b\cdot x=b\cdot x' \\
&\Rightarrow b^{-1}(b\cdot x)=b^{-1}(b\cdot x') \\
&\Rightarrow (b^{-1}b)\cdot x=(b^{-1}b)\cdot x' \\
&\Rightarrow e\cdot x=e\cdot x' \\
&\Rightarrow x=x' \\
\end{alignat}

*

*Surjectivity: $\forall y \in B, \exists x\in A\mid y=b\cdot x \iff \forall y \in B, \space x=b^{-1}\cdot y\in A$; but indeed $a\cdot(b\cdot x)=a\cdot(b\cdot (b^{-1}\cdot y))=a\cdot y\stackrel{y\in B}{=}y=b\cdot x$, and then $x\in A$.

Actually, the finiteness of $X$ doesn't seem to be relevant here.
A: Hint: define a map $\phi : Fix_X(a) \rightarrow Fix_X(b^{-1}ab)$ by $\phi(x)=x^b$. Show that $\phi$ is well-defined and both injective and surjective.
