Suppose that I have $X_i \sim E(1)$,iid, i goes from 1 to n, E stands for exponential distribution, and I want to know the distribution of $\bar{X} = \Sigma_iX_i/n$.
I know that $\Sigma_i X_i \sim Gamma(n,1)$, but how to deal with the remaining 1/n?
Does it affect the shape parameter in the same way it afects scale parameters in location-scale distributions, so that I can say $\bar{X} \sim Gamma(n,n)$?
On
No, it does not work like that. The only thing you can conclude is that $$ n\cdot\overline{X}\sim \text{Gamma}(n,1)$$ which is almost always enough since from here it is not hard to calculate the mean or variance for example.
Edit: I am sorry, you were doing it correctly, also see: How to scale a gamma distribution
(Assuming independence) your $\bar{X} = \Sigma_iX_i/n$ has mean $1$ and variance $\frac1n$ as the average of $n$ random variables with mean and variance $1$
It does have a Gamma distribution, with density $\dfrac{n^n}{\Gamma(n)}x^{n-1}e^{-n x}$
Some people would call this $\text{Gamma}(n,n)$, while others would call it $\text{Gamma}\left(n,\frac1n\right)$ depending on whether they use a rate or a scale parameter as the second term; for the sum rather than the average they would label them the other way round