If $(x_n)_{n=1}^{\infty}$ is periodic then $(x_{f(n)})_{n=1}^{\infty}$ is periodic Let $(x_n)_{n=1}^{\infty} \subset \mathbb{R}$ a periodic sequence, and let $f: \mathbb{N} \to \mathbb{N}$ a polynomial

Is the sequence $(x_{f(n)})_{n=1}^{\infty}$ also periodic?

For example, if we take $(x_n)_{n=1}^{\infty} = (a,b,a,b,a,b, \dots)$ and $f(x) = x^2$ then we get
$$(x_{f(n)})_{n=1}^{\infty}=(a,b,a,b,a,b \dots) = (x_n)_{n=1}^{\infty}$$ since the square preserves the index modulo $2$.
This seems to work for every polynomial. i.e, it preserves the index modulo $2$ for all $n$ or flips it for all $n.$
Is this true for the general case $(x_n)_{n=1}^{\infty} = (a_1,a_2, \dots, a_k,a_1,a_2, \dots a_k, \dots)$? 
 A: Yes. This is relatively easy to prove if the coefficients of the polynomial $f$ are all integers. Then, the process of calculating $f(n)$ only involves integers, their multiplication and addition, which are "preserved" modulo any number, as you noted yourself $\mod 2$.
That means 
$$i \equiv j \pmod k \Rightarrow f(i) \equiv f(j) \pmod k.$$
Since your sequence $(x_n)$ is periodic with period $k$, $(x_{f(n)})$ is also periodic with period $k$.
If the coefficients of $f$ are all rational numbers, then we can find the least common multiple of the coefficients' denominators, say $q>0$, and write
$$f(n) = \frac{g(n)}{q},$$
where $g(n)$ now has only integer coefficients. We then know that 
$$i \equiv j \pmod {qk} \Rightarrow g(i) \equiv g(j) \pmod {qk} \Rightarrow f(i) \equiv f(j) \pmod {k},$$
which proves that $(x_{f(n)})$ has period $qk$.
An example would be $f(n)=\frac{n(n+1)}2$, which starts with $f(1)=1, f(2)=3, f(3)=6, f(4)=10, \ldots$ and then repeats this 2 odd, then 2 even numbers pattern indefinitely. In your example of $(x_n)=(a,b,a,b,\ldots)$ that would mean $(x_{f(n)})=(a,a,b,b,a,a,b,b,\ldots)$.
Now this concludes the proof, as any polynomial $f:\mathbb Z \to \mathbb Z$ must have rational coefficients (not a simple fact, but it can be shown that $n!a_n$ must be an integer if $a_n$ is the leading coefficient of the polynomial and then some induction can made for the next lower coefficients).
I'll also note that the proposition stays true if $(x_n)$ is only periodic after a given index $N$. If $f$ is a constant polynomial, the result is trivial. Otherwise, it must tend to $\infty$ as $n \to \infty$, so after some index $N_0$ the index $f(n)$ will always be in the part where $(x_n)$ is peridic.
A: In this answer, it is shown that any degree-$k$ polynomial $\mathbb{Z}\mapsto\mathbb{Z}$ can be written as
$$
P(n)=\sum_{j=0}^ka_j\binom{n}{j}\tag1
$$
where $a_j\in\mathbb{Z}$. This implies that $k!P(n)\in\mathbb{Z}[n]$, so that if
$$
n_1\equiv n_2\pmod{k!m}\tag2
$$
then
$$
k!P(n_1)\equiv k!P(n_2)\pmod{k!m}\tag3
$$
and therefore,
$$
P(n_1)\equiv P(n_2)\pmod{m}\tag4
$$
Thus, if the period of $x_n$ is $m$, the period of $x_{P(n)}$ is at most $k!m$

Example
The sequence $n$ repeats mod $6$ with period $6$, but the sequence $\binom{n}{3}$ repeats mod $6$ with period $36$:
$$
\scriptsize
0,0,0,1,4,4,2,5,2,0,0,3,4,4,4,5,2,2,0,3,0,4,4,1,2,2,2,3,0,0,4,1,4,2,2,5,\color{#C00}{0,0,0,1,\dots}
$$
