Is this mapping linear? I'm given the following question: 
Consider the operation on a vector that rearranges the components in increasing order. For example: $T(3, 2, 6) = (2, 3, 6)$. Is this a linear mapping? 
Now, can I assume this is NOT a linear mapping because of the following fact: 
$T(3, 2, 6) = T(6, 2, 3) = (2, 3, 6)$? 
 A: No you cannot assume that.
A linear map can map different vectors to the same vector.
Since $(3,2,6)$ and $(6,2,3)$ are linearly independent, there is a linear map which maps them both to $(2,3,6)$
One approach hint: Try to disprove the property that $T(\alpha x) = \alpha T(x)$
Different approach hint: Consider what $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ get mapped to, and see what you can say about the whole map.
A: Consider $T((1,0,0)+(0,1,0))$. 
A: What you've done so far does not suffice to show that $T$ is not linear. 
To show that $T$ is not linear, you have to show that one of the following two properties does not hold:
$\ \ \ \ $1) $T(a{\bf x})= aT({\bf x})$ for all scalars $a$ and all vectors $\bf x$.
$\ \ \ \ $2) $T({\bf x}+ {\bf y})=T({\bf x})+T({\bf y})$ for all vectors $\bf x$ and $\bf y$.

You've noticed $$\tag{ 3 }T(3,2,6)=T(6,2,3)=(2,3,6).$$
Now notice, by the definition of $T$, that $$T\bigl( (3,2,6)+(6,2,3)\bigr)=T(9,4,9)=(4,9,9).$$
But from $(3)$
$$
T(3,2,6)+T(6,2,3)=(2,3,6)+(2,3,6)=(4,6,12)\ne(4,9,9).
$$
So, 2) fails and, thus, $T$ is not linear.
Note that you need only verify that one of the two properties fails. Here, you could have shown that property 1) fails instead:
 $T(1,2,3) =(1,2,3)$ but $$T(-1,-2,-3)=(-3,-2,-1)\ne(-1)T(1,2,3).$$ 
