Forgive my lack of understanding, but I'm looking for a continuous distribution where for some mean $\mu$ and real number $a > 0$, the probability of choosing $\frac{\mu}{a}$ and $a\mu$ is equally likely.
Just as an example, let $\mu = 1$, and $a = 2$. In a regular gaussian graphed here, it's obvious that values chosen will be skewed to be higher than the mean (as there's more area from [1,2] than [0.5,1]). Basically, what I'm looking for is a distribution that fixes this, resulting in the same area from [1,2] as [0.5,1] (for this example, but more generally [$\frac{\mu}{a}$, $\mu$] and [$\mu$, $a\mu$]).
I feel like this has to be a thing, I just haven't worked much with probability, and don't know exactly what to google.
Best Answer
You want a random variable $X$ such that
$$\int_{\mu/a}^{\mu}f_X(x)\;\mathrm{d}x=\int_{\mu}^{\mu a}f_X(x)\;\mathrm{d}x$$
If we make the change of variable $x=e^u$, (with $\mathrm{d}x=e^u\mathrm{d}u$), we have
$$\int_{\log \mu - \log a}^{\log\mu}f_X(e^u)e^u\;\mathrm{d}u=\int_{\log\mu}^{\log \mu + \log a}f_X(e^u)e^u\;\mathrm{d}u$$
This must hold for all $a$, so $f_U(u)=f_X(e^u)e^u$ must be symmetric about $\log\mu$. Therefore, what you need to do is:
Pick any continuous distribution $f_U(u)$ that is symmetric about $\log\mu$.
Compute a new distribution $f_X(x)=\frac{f_U(\log x)}{x}$. This is, as kimchi lover's comment says, the probability density function of $X=e^U$.
The distribution $f_X(x)$ satisfies your criterion.
For example, if you apply this procedure starting with a Gaussian distribution, you get the log-normal distribution, which satisfies your criterion (note that the Wikipedia article defines $\mu$ differently: you need to use $\log\mu$ here):
$$f_X(x)=\frac{1}{x\sigma\sqrt{2\pi}}e^{-\frac{\left(\log x - \log \mu\right)^2}{2\sigma^2}}.$$
Edit: this method produces random variables $X$ that satisfy
$$\mathrm{Pr}\left[X\in\left[\frac{\mu}{a}, \mu\right]\right]=\mathrm{Pr}\left[X\in[\mu, \mu a]\right]\qquad\quad(1)$$
for a fixed $\mu$ and any $a$, but it does not make the mean $\mathrm{E}[X]$ equal to $\mu$. We have shown that, for (1) to hold, it is required that $U=\log X$ is symmetric about $\log\mu$. As pointed out by guy in the comments, this means that $\mathrm{E}[U]=\log\mu$, and Jensen's inequality tells us that
$$\mathrm{E}[X]=\mathrm{E}[e^U]\ge e^{\mathrm{E}[U]}=\mu.\qquad\quad(2)$$
As $e^U$ is a strictly convex function, equality holds only when $U$ is constant. Therefore, the only distribution that satisfies (1) and has mean $\mu$ is the constant distribution
$$\textrm{Pr}[X=\mu]=1.$$