For the second problem, we can cheat, and let $X=-1$ with probability $p$, and $X=1$ with probability $1-p$. The random variables $X$ and $\dfrac{1}{X}$ not only have the same distribution, they are the same random variable.
Now that we are cheating, let's go all the way. Let $X=1$ with probability $1$.
For cheating a little less, let $X$ take on the values $\dfrac{1}{2}$ and $2$ each with probability $\frac{1}{2}$. Then $X$ and $\dfrac{1}{X}$ are not the same random variable, but they have the same distribution.
A continuous distribution is more challenging. But for example a random variable $X$ with density function $f_X(x)=\dfrac{1}{x}$ for $e^{-1/2}\le x\le e^{1/2}$ and $f_X(x)=0$ elsewhere has the desired property.