When dealing with this kind of functions, you should divide the range of function so that you could eliminate canonical forms then prove that they are identical in fact. For demonstration, let's prove that your $f,g,i$ are same by dividing the range of function.
Note that the parameters of $\max$s and $\min$s in your $g,h,i$ are integer multiples of $x$ or their $\max$, $\min$ result. Therefore, we only need to consider two cases: $x \ge 0$ and $x<0$.
i) $x \ge 0$
$2x \ge x \ge 0 \ge -x \ge -2x \ge -3x$. Therefore
$$g(x)=\max(-x,x)=x\\h(x)=\max(\min(-x,-2x),\min(x,2x))=\max(-2x,x)=x\\i(x)=\max(\min(-x,-3x),x)=\max(-3x,x)=x$$
ii) $x < 0$
$2x < x < 0 < -x < -2x < -3x$. Therefore
$$g(x)=\max(-x,x)=-x\\h(x)=\max(\min(-x,-2x),\min(x,2x))=\max(-x,2x)=-x\\i(x)=\max(\min(-x,-3x),x)=\max(-x,x)=-x$$
By i) and ii), $g,h,i$ are identical.
No, they are not.
For instance, $2x^2-2$ and $x^2-1$ have the same roots, yet they are not identical.
And, depending on what you mean by "the same roots", we have that $x^2-2x+1$ and $x-1$ have the same roots, yet they are not identical.
Again, depending on what you mean by "the same roots", $x^3+x$ and $x^3+2x$ both only have one real root, yet they are not the same.
However, if two monic polynomials have the same roots, with the same multiplicities, over some algebraicaly closed field (like the complex numbers $\Bbb C$) then yes, they are identical.
Best Answer
If $f$ and $g$ are two polynomials of degree $n$, and there are distinct points $x_1, \dots, x_{n+1}$ such that $f(x_i) = g(x_i)$ for $i = 1, \dots, n+1$, then $f = g$. One way to see this is that a generic degree $n$ polynomial has the form
$$a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$
which contains $n+1$ coefficients. Given $n+1$ points $(x_i, y_i)$ which satisfy the polynomial equation, we obtain $n+1$ equations in $n+1$ unknowns which uniquely determine the coefficients, and hence the polynomial.
If now $f$ and $g$ are polynomials in two variables of degree $n$. A generic such polynomial has the form
$$\sum_{i=0}^n\sum_{j=0}^ia_{ij}x^jy^{i-j}$$
which contains $N:=\frac{(n+1)(n+2)}{2}$ coefficients. Given the situation for one variable polynomials, you may guess that if there are distinct points $(x_1, y_1), \dots, (x_N, y_N)$ with $f(x_i, y_i) = g(x_i, y_i)$ for $i = 1, \dots, N$, then $f = g$. Your guess would be wrong. While it is true that $N$ points of agreement is enough to show that $f = g$, it is not true that any $N$ points will do. Each point will still give an equation so that you obtain $N$ equations in $N$ unknowns, you might get some redundancy which means the system will not have a unique solution.