Explanation about the shape of an object of a linear transformation. We have the linear mapping $A: R^2 \to R^2$. In which case (see picture) we can conclude that $A$ might be a linear mapping?
a.) and d.) are not correct because we can immediately see that the zero vector doesn't map to the zero vector. But how can we choose between b.) , c.) and e.)? Does a linear mapping preserve the shape of the object?

You have correctly said that (a) and (d) are not possible, because nothing maps to $(0,0)$.

Let us start by showing that (c) works. It should not be very difficult to see that if we take the map
$$\mathcal A \colon (x,y) \mapsto \left(\frac{x+y}2,\frac{x+y}2\right)$$
we get the desired result.
This map maps $(0,0)\mapsto(0,0)$. Both $(0,1)$ and $(1,0)$ are mapped to $(\frac12,\frac12)$. The point $(1,1)$ is mapped to $(1,1)$.
Try to check in detail that you get all points of the line segment, and no other points.

Now we want to show that there is no linear map $\mathcal A$ which works as in (b) or (e).
Maybe symmetry might help us. The original figure $H$ is symmetric w.r.t the point $(\frac12,\frac12)$. This means that
$$\left(\frac12,\frac12\right)+(x,y)\in H \implies \left(\frac12,\frac12\right)-(x,y)\in H.$$
Then the image will be symmetric w.r.t. the point $\mathcal A(\frac12,\frac12)$. Indeed, we have
$$\mathcal A\left(\frac12,\frac12\right)+\mathcal A(x,y)\in \mathcal A[H] \implies \mathcal A\left(\frac12,\frac12\right)-\mathcal A(x,y)\in \mathcal A[H].$$
But if we look at the pictures, neither the figure in (b) nor the figure in (e) has a center of symmetry. 
