Time shifts and change in phase
For a continuous-time signal $x(t)$ with Fourier Transform $X(f)$, the 
Fourier transform of $x(t-t_0)$ is $e^{-j2\pi ft_0}X(f)$, that is, the value
of $X(3)$, say, gets changed to $e^{-j2\pi 3t_0}X(3)$ which has the same
magnitude as $X(3)$ but a different phase.

For a discrete-time continuous-amplitude signal $x[n]$ with DFT $X[k]$
of length $N$, you have essentially discarded every piece of information
conveyed to you by $x[-1], x[-2], \ldots $ and by $x[N], x[N+1], x[N+2], \ldots$
and have retained only $x[0], x[1], \ldots, x[N-1]$. If you phase-shift
the DFT appropriately, then yes, you will get a circular shift of the $x[n]$
values. The result will be the linear shift of the $x[n]$ values only
in the fortuitous circumstance that the sequence $x[n]$ is itself periodic
with period $N$, and so the untrained observer such as myself cannot tell
if that symbol that just shifted into my DFT wndow is $x[-1]$ coming from
the left (linear shift) or $x[N-1]$ from the right (circular shift)
because by golly they look exactly the same to me.  Savvier denizens
of this forum will give you more elaborate answers explaining how you
can tell the difference between the two cases.