= Solution
If $A$ is nonempty, take $a\in A$. The stated implication repeatedly gives $a+jt\in A$ for $j=0,1,\ldots,p-1$. These residues are all distinct: equality of two would give $(j-\ell)t=0$ in the <prime field> $\mathbb Z_p$, and the nonzero $t$ can be cancelled. They therefore exhaust the field. Thus
$$
\boxed{A+t\subseteq A,\quad t\ne0\quad\Longrightarrow\quad A=\varnothing\text{ or }A=\mathbb Z_p.}
$$
Equivalently, a nonzero element generates the additive <cyclic group> of prime order.
Now assume $0<k<p$. If $A+s=A+t$ with $s\ne t$, translating by $-s$ gives $A=A+(t-s)$, contradicting the preceding result. Hence all $p$ translates are distinct. Translation defines equivalence classes on the collection of all $k$-element subsets: two subsets are equivalent if one is a translate of the other, and each class contains exactly $p$ subsets. Since there are $\binom pk$ such subsets, their partition into classes gives
$$
\boxed{p\mid\binom pk\qquad(1\leq k\leq p-1).}
$$
This is the <free translation action on nontrivial subsets of a prime cyclic group>.
The <binomial theorem> now gives, in the <prime field>,
$$
(a+1)^p=a^p+1+\sum_{j=1}^{p-1}\binom pj a^j= a^p+1.
$$
Starting with $0^p=0$ and applying this identity successively to the representatives $0,1,\ldots,p-1$ proves $a^p=a$ for every residue $a$. For nonzero $a$, cancellation gives
$$
\boxed{a^p=a\text{ in }\mathbb Z_p;\qquad a^{p-1}=1\text{ if }a\ne0,}
$$
which is <Fermat's little theorem>, or $a^p\equiv a\pmod p$ for every integer $a$ and $a^{p-1}\equiv1$ when $p\nmid a$.
The same vanishing of intermediate <binomial coefficients> proves $(R+S)^p=R^p+S^p$ for any commuting elements in a ring whose <ring characteristic> is $p$, in particular for polynomials over $\mathbb Z_p$. Applying this repeatedly to the terms of $Q$, and applying <Fermat's little theorem> to each coefficient, gives the <Frobenius endomorphism> identity
$$
Q(x)^p=\sum_{j=0}^n(a_jx^j)^p=\sum_{j=0}^na_j^px^{pj},\qquad \boxed{Q(x)^p=\sum_{j=0}^na_jx^{pj}=Q(x^p).}
$$
This is an equality of formal <polynomials>. It does not assert $x^p=x$ as a polynomial; that latter equality holds only after evaluating at elements of $\mathbb Z_p$.
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