The function $f(x)$ satisfies
\[f(xy) = f(x) f(y)\]for all real numbers $x$ and $y.$ Find all possible values of $f(1).$  Enter all the possible values, separated by commas.
Setting $x = y = 1,$ we get
\[f(1) = f(1)^2,\]so $f(1) = 0$ or $f(1) = 1.$

The function $f(x) = 0$ shows that $f(1)$ can be 0, and the function $f(x) = x$ (and the function $f(x) = 1$) shows that $f(1)$ can be 1.  Thus, the possible values of $f(1)$ are $\boxed{0,1}.$