Delayed impulse does not delay spectral response
The convolution theorem ("Multiplication in time is Convolution in frequency") states:

$$\mathcal{F}\{x\cdot y\}=\mathcal{F}\{x\}*\mathcal{F}\{y\}$$

where $x, y$ are two time-domain signals, $\cdot$ denotes element-wise multiplication and $*$ is the convolution operator. So, the convolution in the frequency domain, is carried out between the Fourier Transforms of the original signals. 

As an example, let us use your impulses with frequency response:

$$\mathcal{F}\{\delta(t)\}=1$$

Further, note the time-shifting property of the FT:

$$\mathcal{F}\{x(t-t_0)\}=\mathcal{F}\{x(t)\}\exp(-j2\pi ft_0)$$

where $x(t)$ is any signal you like.
Now, you can readily explain the results you obtain for your shifted impulses:

$$\mathcal{F}\{\delta(t-1)\}=\mathcal{F}\{\delta(t))\}\exp(j2\pi f)$$

where $f=[0/N, 1/N, 2/N, \ldots, (N-1)/N]$ (due to the discrete FT).  Similar for the two-shifted case:

$$\mathcal{F}\{\delta(t-2)\}=\mathcal{F}\{\delta(t)\}\exp(-j2\pi f2)$$

Note that $\mathcal{F}\{\delta(t)\}=1$. You display the absolute value of your spectra: $\left|X(f)\exp(j2\pi ft_0)\right|=|X(f)|$ for any $X(f)$ you like. That's why you see three times the same output. Try plotting real and imaginary part separately, then you will see the effect of time-shifting. You might also have a look here which is an article I wrote about the effect of the time-shift and frequency-shift.