If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y_0^2,\ldots,Y_n^2$? I don't have a very solid foundation in measure theory, and this always seems a bit confusing to me so I would appreciate any help.
We are given 
$
E \left( X_{n+1} | Y_0,\ldots,Y_n \right) = X_n.
$
Prove or disprove
$
E \left( X_{n+1} | Y_0^2,\ldots,Y_n^2 \right) = X_n
$
I am thinking that if $F=\sigma \left(Y_0,\ldots,Y_n \right)$ and
$G=\sigma \left(Y_0^2,\ldots,Y_n^2 \right)$, I need to prove that F = G? Is this correct?
Then I can do something like this:
$
E \left( X_{n+1} | G \right) = E \left( E \left( X_{n+1} | F \right) | G \right) = E \left( X_{n+1} | F \right) = X_{n+1}
$.
Also is $E \left( X_{n+1}^2 | G \right) = X_n^2$ (a martingale) given
$
E \left( X_{n+1} | Y_0,\ldots,Y_n \right) = X_n.
$
 A: 
Prove or disprove
  $
E \left( X_{n+1} | Y_0^2,\ldots,Y_n^2 \right) = X_n
$
I am thinking that if $F=\sigma \left(Y_0,\ldots,Y_n \right)$ and
  $G=\sigma \left(Y_0^2,\ldots,Y_n^2 \right)$, I need to prove that F = G? Is this correct?

Actually, $G\subseteq F$ and the inclusion can be strict.

Then I can do something like this: $E \left( X_{n+1} | G \right) = E \left( E \left( X_{n+1} | F \right) | G \right) = E \left( X_{n+1} | F \right) = X_{n+1}
$.

The third equality is wrong. When $G\subseteq F$, $E \left( E \left( X_{n+1} | F \right) | G \right)= E \left( X_{n+1} | G \right) \ne E \left( X_{n+1} | F \right)$ in general hence this proves nothing.
Here is a counterexample to the statement you are trying to show: if $(Y_k)$ is an i.i.d. Bernoulli sequence, then $Y_n^2=1$ almost surely hence $E \left( X_{n+1} | Y_0^2,\ldots,Y_n^2 \right) = E(X_{n+1})\ne X_n$ in general.

Also is $E \left( X_{n+1}^2 | G \right) = X_n^2$ (a martingale) given
  $
E \left( X_{n+1} | Y_0,\ldots,Y_n \right) = X_n.
$

No. Actually, by convexity, $E \left( X_{n+1}^2 | G \right) \geqslant E \left( X_{n+1}| G \right)^2= X_n^2$ and the equality holds if and only if $X_{n+1}$ is measurable with respect to $G$.
A: $G \subseteq F$
If we know $Y_0, Y_1, ...$, then we know $Y_0^2, Y_1^2, ...$ The converse is not true.
$E[X|G] = E[E[X|F]|G]$, but $E[X|F] \ne E[E[X|G]|F]$.
$E[X|G] = E[E[X|F]|G]$ allows you to prove the converse of your conjecture is true.
