Prove that $\mathrm{Ind}_{H}^{G}(U) \cong \mathbb{C}X$ as $G$-reps where $X$ is transitive, $H = G_x$, $U$ is the trivial $H$-rep 
Let $G$ be a finite group, let $X$ be a transitive $G$-set and let $V = \mathbb{C}X$ be the corresponding permutation representation of $G$. Let $H = G_x$ be the stabiliser of an element $x \in X$ and let $U$ be the trivial representation of $H$.
Prove that $\operatorname{Ind}_{H}^{G}(U)$ and $\mathbb{C}X$ are isomorphic representations of $G$.

So my initial thoughts are to use that
$$
  \operatorname{Ind}_H^G(\chi(g))
  =
  \frac{1}{|H|} \sum_{x \in G} \chi^0( xgx^{-1} )
$$
where
$$
  \chi^0
  =
  \begin{cases}
    1 & \text{if $xgx^{-1} \in H$}, \\
    0 & \text{if $xgx^{-1} \notin H$}
  \end{cases}
$$
(where $\chi$ is the character of the trivial representation), and then use the fact that
$$
  \psi_{\mathbb{C}X}(g)
  = \operatorname{fix}(g)
  = |\{ x \in X: g \mathbin{.} x = x \}| \,.
$$
And then show that these two are equal, but I'm totally stuck on how to do that.
 A: Let me fix $x_0 \in X$ such that $H = G_{x_0}$ is the stabilizer of $x_0$, and let me fix $g\in G$. So that we do not confuse it with an element of $X$, let me change the variable of your sum from $x$ to $a$. Also, via the change of variable $a \mapsto a^{-1}$, I write it the following way (for convenience)
$$Ind_H^G\left(\chi_{ }\left(g\right)\right)=\frac{1}{\left|H\right|}\sum_{a\in G}^{ }\chi^0\left(a^{-1}ga\right).$$
By definition of $\chi^0$, we have $\chi^0(a^{-1}ga) = 1 \iff g\in aHa^{-1} = G_{a\cdot x_0}$, ie. if and only if $g$ fixes $a\cdot x_0$. Otherwise, $\chi^0(a^{-1}ga)$ is zero. Therefore, we have
$$Ind_H^G\left(\chi_{ }\left(g\right)\right)=\frac{1}{\left|H\right|}\#\{a\in G \,|\, g\in G_{a\cdot x_0}\}.$$
Let us consider the composite map (of sets) $\varphi: G \rightarrow G/H \xrightarrow{\sim} G\cdot x_0 \subset X$, where the first arrow is the quotient map and the second arrow is the usual bijection between $G/H$ and the orbit of $x_0$ defined by $aH \mapsto a\cdot x_0$. I claim that the restriction of $\varphi$ defines a surjection
$$\varphi:\{a\in G \,|\, g\in G_{a\cdot x_0}\} \to \left\{x\in X \,|\, \ g.x=x\right\},$$
and that each fiber of this surjection has cardinality $|H|$. Indeed, if $a$ belongs the the LHS, we have $\varphi(a) = a\cdot x_0$, which belongs to the RHS. Thus the map above is well defined (ie. the restriction of $\varphi$ to the LHS really falls into the RHS).
Moreover it is surjective. For this, let $x$ be in the RHS. By transitivity of the $G$-action on $X$, there exists some $a\in G$ such that $x = a\cdot x_0$. Since $g\cdot x = x$, it follows that $a$ belongs to the LHS and we have $x = \varphi(a)$.
Eventually, let us look at the fibers. Let $x$ be in the RHS and write $x = a\cdot x_0$ for some $a\in G$ as before. Let $a'$ be another element such that $x = a'\cdot x_0$. Then $a\cdot x_0 = a'\cdot x_0$ which means that $a^{-1}a' \in H$. Therefore $a' \in aH$. Conversely, one may check that all elements of $aH$ belong to the LHS and are contained in the fiber of $x$. Hence, we do have $\varphi^{-1}\{x\} = aH$.
Therefore, it follows that
$$\#\{a\in G \,|\, g\in G_{a\cdot x_0}\} = \# H \# \left\{x\in X \,|\, \ g.x=x\right\},$$
which concludes the proof.
