Suppose X and Y are i.i.d and $\frac{X+Y}{\sqrt{2}}\overset{d}=X\overset{d}=Y$, show that X has a normal distribution. Suppose $X$ and $Y$ are independent with mean zero and variance 1, and
$\frac{X+Y}{\sqrt{2}}\overset{d}=X\overset{d}=Y$. Use the CLT to show that $X\sim \mathcal N(0,1)$.
I saw in many places that set $\frac{S_{2^n}}{\sqrt{2^n}}$ to prove the problem. Is it feasible to go through the proof by using $\frac{S_{2n}}{\sqrt{2n}}$?

I see, we need to set $\frac{S_{2^n}}{\sqrt{2^n}}$. Thus $X\overset{d}=\frac{S_{2^n}}{\sqrt{2^n}}\overset{d}\to \mathcal N(0,1)$.
 A: Using the line of reasoning suggested by the author, let 
$$S_{2^n}=\frac{1}{\sqrt2^{n}}(X_1+...+X_{2^n}),\ n=1,2,3,...$$
where $X_i$ are independent draws from the distribution of $X$. Now 
$$S_{2^{n}}=\frac{1}{\sqrt2^{n-1}}(\frac{X_1+X_2}{\sqrt{2}}+...+\frac{X_{2^{n}-1}+X_{2^n}}{\sqrt{2}}),\ n=1,2,3,...$$
But you have $S_{2^{n}} \overset{d}= S_{2^{n-1}}$. Repeating this iteratively $X \overset{d}= S_{2^{n}} \overset{d}\to \mathcal{N}(0,1)$, where the last result follows from CLT.
A: Here is a bit tedious solution using characteristic functions.
The characteristic function of a sum of independet variables
$$
\varphi_{(X+Y)/\sqrt2}(t)=\varphi_X(t/\sqrt2)\varphi_Y(t/\sqrt2),
$$
and since all three distributions are the same,
$$
\varphi_X(t)=\varphi_X^2(t/\sqrt2).
$$
Since the first two moments of $X$ exist, the function $\varphi_X(t)$ is continuously differentiable twice, its log $\psi_X(t)=\log\varphi_X(t)$ is continuously differentiable twice too.
For the $\psi_X(t)$ and its derivatives we have the equations
\begin{align*}
\psi_X(t)&=2\psi_X(t/\sqrt2),\\
\psi_X'(t)&=\sqrt2\psi_X'(t/\sqrt2),\\
\psi_X''(t)&=\psi_X''(t/\sqrt2).
\end{align*}
Note that the second derivative is continuous at zero, so it is constant (if it takes two different values at $t_1$ and $t_2$ then it takes the same values at the sequences $t_1/(\sqrt2)^n$ and $t_2/(\sqrt2)^n$ and have a discontinuity in zero).
Integrating twice we have the function $\psi_X(t)=a+bt+ct^2$ and $\varphi_X(t)=e^{a+bt+ct^2}$. Note that $a=0$ (otherwise $\varphi_X(0)\ne0$), and $b=0$ (because $E(X)=0$). Also, since $E(X^2)=-\varphi_X''(0)=-2c=1$, we have $c=-1/2$, and finally
$$
\varphi_X(t)=e^{-t^2/2}
$$
which is a characteristic function of the standard normal distribution.
