How do I transform the parameters of the F distribution? Motivation
I have a prior on a random variable $X\sim \beta^\prime(\alpha,\beta)$ that I would like to use in JAGS. JAGS does not support the $\beta^\prime$ distribution, but it does support the $F$ distributions, and the F distribution is related to the $\beta^\prime(\alpha,\beta)$ like this:
$$\text{if } X\sim \beta^\prime(\alpha,\beta) \text{ then } X\frac{\alpha}{\beta}\sim F(2\alpha, 2\beta)$$
Question 
Is there a transformation $$c,d = f(\alpha, \beta)$$ such that:
$$\text{if } X\sim \beta^\prime(\alpha,\beta)\text{ then } X\sim F(c, d)?$$ 
What is an appropriate analytical approach or solution to this problem?
Current Approach
My solution is to use simulation. Although sufficient for my application, a formal solution would be more satisfying.
   set.seed(0)
   alpha <- 2
   beta  <- 4
   Y <- rf(100000, 2*alpha, 2*beta) * ( beta / alpha )
   parms <-signif(fitdistr(Y, 'f', start = list(df1=1, df2=2))$estimate,2)

Update whuber's answer states that there is no general transformation.
 A: I don't quite get it: why can't you just generate an $F_{2\alpha, 2\beta}$ variate and rescale it by $\beta/\alpha$?  I also don't see the connection between the code in your current approach and your question, because $Y$ appears to be an $F_{2\alpha, 2\beta}$ variate rescaled by $\alpha/\beta$ rather than  $\beta/\alpha$.
Nevertheless, taking the question at face value, if there are parameters $c$ and $d$ corresponding to $\alpha$ and $\beta$, then


*

*The mean of $X$ must be $\beta/\alpha$ times the mean of $F_{2\alpha, 2\beta}$ and it is also equal to the mean of $F_{c,d}$.

*The mode of $X$ must be $\beta/\alpha$ times the mode of $F_{2\alpha, 2\beta}$ and it is also the mode of  $F_{c,d}$.
Using standard formulas and solving gives a unique solution
$$\eqalign{
c = & \frac{2 \left(\alpha^2+\alpha^2 \beta \right)}{\alpha -\alpha  \beta + 2 \alpha^2 \beta + 2 \beta ^2-2 \alpha  \beta^2} \cr

d = & \frac{2 \beta^2}{\alpha -\alpha \beta +\beta^2}.
}$$
We can also equate variances:

*The variance of $X$ must equal $(\beta/\alpha)^2$ times the variance of $F_{2\alpha, 2\beta}$ and it must equal the variance of $F_{c,d}$.
Except possibly for special values of $\alpha$ and $\beta$ (most notably, $\alpha = \beta$), this requirement conflicts with the preceding solution.  Therefore there does not exist any such transformation in general.
