EDIT: Corrected an error in my OP.
Let $X,Y$ be independent random variables of the same "type" or "class" (see examples below), and let $Z=X+Y$. I am interested in the conditional distribution of $Y$, conditioned on $Z=z$, which I will denote as $(Y|Z=z)$. I noticed this:
If $X,Y$ were independent uniform random variables, e.g. $U(x_1,x_2), U(y_1, y_2)$, then clearly for any $z, (Y|Z=z)$ is also $\sim U(a,b)$ for some $a$ and $b$ which depend on $z$, because conditioning on $Z=z$ means restricting to a $-45°$ line within the $[x_1, x_2] \times [y_1, y_2]$ rectangle. (Note that $Z$ itself is not uniform, but triangular distributed.)
If $X \sim N(\mu_x, \sigma^2_x), Y \sim N(\mu_y, \sigma^2_y)$ are independent Gaussians, then I believe $Y,Z$ are multivariate Gaussian (although no longer independent) and consequently $(Y|Z=z)$ (for any $z$) is still Gaussian but with different parameters (mean and sigma) that depend on $z$. (I am not 100% sure about this, and my question on this specific case is a different post: Gaussians and sum of Gaussians)
So we have two examples where, $X$ and $Y$ both belong to a certain "class" (Gaussian or Uniform), and conditioned on $Z=X+Y=z$, the marginal $(Y|Z=z)$ still belong to the same class (Gaussian or Uniform), for any $z$. Sorry I don't have a very precise definition of "class" but how general is this phenomenon? Or is there something special about Uniform (based on simple geometry) or Gaussian (based on complicated proofs about properties of multivariate Gaussians)? Note that $Z=X+Y$ does not necessarily belong to the same class (it does in case of Gaussian, but not in case of Uniform).
There are many counter-examples, and the proofs are usually simply "range" violations, such as:
If $X, Y$ are independent $Bin(n,p)$, then $(Y|Z=z)$ is not binomial for some $z$. E.g. $z = n+1 \Rightarrow Y \ge 1$ (because $X \le n$), but no binomial r.v. (with $p < 1$) can have a range that exclude $0$. (However I am unsure if $(Y|Z=z)$ in this case is a shifted binomial, i.e. some $Bin(n', p') +$ some constant.)
More generally, if the "class" consists of r.v.s with range $[0, +\infty)$, then $(Y|Z=z)$ no longer belong to that class since $(Y|Z=z) \le z$ and does not have the required range to infinity. This rules out: Poisson, exponentials, and many other classes.
So, besides Uniform and Gaussians, are there other such "classes" where if $X, Y$ belong to the class then $(Y|Z=z)$ also belong to the class, for any $z$? And are there underlying properties that make this possible?
Here is a counter-example. Suppose $X$ and $Y$ are exponential: $\Pr(X>t) = \Pr(Y>t) = e^{-t}$ for $t\ge 0$. Then $\Pr(Y>t | X+Y=z) = 0$ if $t > z > 0$.