There are $p$ solutions to this question, one for each value from $0$ to $p-1$. Indeed, it is related to the issue that numbers ending in ...109376 have all their powers ending the same.
The case for $x^p=x \pmod{p^n}$ corresonds to $x^n \mid x^{p-1}$, and it is this second example that we shall look at. In essence, what we shall show, is that the last $n$ digits of $x$ in base $p$, must fall in a particular set of $p$ solutions, one for each digit.
Consider the general number $w, x$ where these are the last two places in base p. Raising this to the power of $p$, will give all multiples of $p^2$, + $pwx^{p-1}, x^p$. But we see already that the 'second remainder' or 'tens column' is already a multiple of $p$, by the binomial expansion, so the last digits of $w, x$, is a particular $x_1, x$.
This can be repeated by replacing $p$ with $p^2$, and shows that for any $x_0$, there is a unique sequence $x_7,x_6,x_5,x_4,x_3,x_2,x_1,x_0$ for every $x_0$, for which this sequence is unique, and must exist.
The missing case is $0$. Of course, one can have sequences ending in $0$, such as $1,0$. But the only instance where the condition holds true here is $p^n\mid x$, which is the case that $0$ is a soultion.
This last point raises questions about the vary nature of primes, since the effect of multiplication by primes seek to make the set more zero-like (even in $\mathbb Z$, so the net result is that one has to add a certain number of non-zero divisors to get 0, and where the case where p is composite, this means that the general endings 'converge' onto a smaller range of values.