Let $X_1, X_2, ..., X_k$ be an independent and identically distributed random variables. Assume $\mathbb{E}(X_i^2) < \infty$ for $1 \leq i \leq k$ and $$\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}} \ \mbox{and} \ X_1$$ has the same distribution function. Then $$X_1 \ \mbox{is normal with mean} \ 0.$$
I notice that the statement is similar to CLT version with iid, but CLT always works with an infinite sequence of random variables.
Here is what I try :
Since $$\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}} =^{distribution} X_1$$ $$\phi_{\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}}}(t) = \phi_{X_1}$$ where $\phi_X(t) = \mathbb{E}(e^{itX})$ the characteristic function.
By independent and identically distributed, $$(\phi_{\frac{X_1}{\sqrt{k}}}(t))^k=\phi_{\frac{X_1 + X_2 + ... + X_k}{\sqrt{k}}}(t) = \phi_{X_1}(t).$$
That is where I do not know how to go on. Any hints please ?
Since the question is tagged "central limit theorem", let me give a way using this result.
We assume $k\geqslant 2$. Consider $\left(Y_i\right)_{i\geqslant 1}$ a sequence of i.i.d. random variables distributed as $X$. By induction on $l$, it is possible to deduce that for each $l\geqslant 1$, $$\frac 1{\sqrt{k^l}}\sum_{i=1}^{k^l}Y_i\mbox{ as the same distribution as }X.$$ Now, since $X$ is necessarily centered and has a finite second moment, we conclude by the central limit theorem that $X$ is a centered (possibly degenerated) Gaussian random variable.