Let $j\ge 1$ be an integer and $p$ prime.
Is it true that
$$\binom{jp}{j}\equiv j\binom{p}{j}\pmod{p^2}$$
My work No, take $j>p$, then the RHS is zero, while the LHS need not be $\equiv 0$. For example, for $p=3$ and $j=5$, I find $\binom{jp}{j}=3003\equiv 6\pmod{p^2}$.
However, is the claim true for $j\le p$?
$\binom{jp}{j} = \frac{jp}{j} \binom{jp-1}{j-1} = p \binom{jp-1}{j-1}$ and $j \binom{p}{j} = j \frac{p}{j} \binom{p-1}{j-1} = p \binom{p-1}{j-1}$. Now we want to show that $\binom{jp-1}{j-1} \equiv \binom{p-1}{j-1} \pmod{p}$.
Since $jp-i \equiv p-i \pmod{p}$, we have $$\binom{jp-1}{j-1} = \frac{(jp-1)\cdots(jp-j+1)}{(j-1)!} \equiv \frac{(p-1) \cdots (p-j+1)}{(j-1)!} = \binom{p-1}{j-1} \pmod{p}$$
Therefore $$\binom{jp}{j} = p \binom{jp-1}{j-1} \equiv p \binom{p-1}{j-1} = j \binom{p}{j} \pmod{p^2}. $$
Comment. $j \le p$ is used in the line expanding $\binom{jp-1}{j-1}$ because it makes sense only when the denominator is nonzero.