If X/Y has the same distribution as Z, is it true that X has the same distribution as YZ? Let X, Y and Z be three independent random variables. If X/Y has the same distribution as Z, is it true that X has the same distribution as YZ? 
 A: It can happen. For instance, if $X$, $Y$ and $Z$ are independent Rademacher variables, i.e. they can be 1 or -1 with equal probability. In this case $X/Y$ is also Rademacher, so has the same distribution as $Z$, while $YZ$ is Rademacher so has the same distribution as $X$.
But it won't happen in general. So long as the means exist, necessary (but not sufficient) conditions for $X/Y$ to have the same distribution as $Z$, and for $YZ$ to have the same distribution as $X$, would be:
$$\mathbb{E}(Z) = \mathbb{E}(XY^{-1}) = \mathbb{E}(X)\mathbb{E}(Y^{-1})$$
$$\mathbb{E}(X) = \mathbb{E}(YZ) = \mathbb{E}(Y)\mathbb{E}(Z)$$
The second equalities followed by independence. Substituting gives:
$$\mathbb{E}(Z) = \mathbb{E}(Y) \mathbb{E}(Z) \mathbb{E}(Y^{-1})$$
If $\mathbb{E}(Z) \neq 0$ then $1 = \mathbb{E}(Y) \mathbb{E}(Y^{-1})$, or equivalently, so long as $\mathbb{E}(Y) \neq 0$,
$$\mathbb{E}(Y^{-1}) = \frac{1}{\mathbb{E}(Y)}$$
This is not true in general. For example, let $Y$ be a translated Bernouilli variable which takes values $1$ or $2$ with equal probability, so $\mathbb{E}(Y)=1.5$. Then $Y^{-1}$ takes values $1$ or $0.5$ with equal probability, so $\mathbb{E}(Y^{-1})=0.75 \neq 1.5^{-1}$. (I leave it to the reader's imagination, how dramatic an effect it would have had to use an untranslated Bernouilli variable instead, or one translated only slightly so it is very close to 0 with probability one half. Note that in the Rademacher example there was no problem here because all three expectations were zero, note further that this condition isn't a sufficient one.)
We can explore how this $Y$ fails by constructing a more explicit counterexample. To keep things simple, suppose $X$ is a scaled Bernouilli and takes values $0$ or $2$ with equal probability. Then $X/Y$ is either $0/1$, $0/2$, $2/1$ or $2/2$ with equal probability. It's clear that $P(X/Y=0)=\frac{1}{2}$, $P(X/Y=1)=\frac{1}{4}$ and $P(X/Y=2)=\frac{1}{4}$. Let $Z$ be an independent variable drawn from the same distribution. What is the distribution of $YZ$? Is it the same as the distribution of $X$? We don't even have to work out the full probability distribution to see that it can't be; it suffices to remember $X$ could only be zero or two while $YZ$ can take any value you can obtain from multiplying one of $\{1,2\}$ by one of $\{0,1,2\}$. 
If you want a moral for this tale, then try playing around with scaled and translated Bernouilli variables (which includes Rademacher variables). They can be a simple way to construct examples - and counterexamples. It helps having fewer values in the supports so that distributions of various functions of the variables can be easily worked out by hand.
Even more extreme we can consider degenerate variables which only have a single value in their support. If $X$ and $Y$ are degenerate (with $Y\neq 0$) then $Z=X/Y$ will be too, and so the distribution of $YZ$ will match the value of $Z$. Like my Rademacher example, that's a situation showing your conditions can be satisfied. If instead, as @whuber suggests in the comments, we let $X$ be degenerate with $P(X=1)$, but allow $Y$ to vary, then constructing an even simpler counterexample is very easy. If $Y$ can take two finite, non-zero values - $a$ and $b$, say - with positive probability, then $X/Y$, and hence $Z$, can take values $a^{-1}$ and $b^{-1}$. Now $YZ$ therefore has $ab^{-1} \neq 1$ in its support, so can't follow the same distribution as $X$. This is similar to, but simpler than, my argument that the supports couldn't match in my original counterexample.
