Which of these signals have same Nyquist rate as $x(t)$? The options are: 


$x^2(t)$
$x(2t)$
Time derivative of $x(t)$
Convolution of $x(t)$ with itself


I guess, the Nyquist rate remains the same as long as bandwidth of $X(w)$ remains the same, as Nyquist rate = 2 * bandwidth.

The Fourier transform of the fourth option is $X^2(w)$. I think this has the same bandwidth as $X{w}$, because the zero terms outside of the bandwidth give zero squares. So this is correct

The third option: Derivative leads to multiplication by jw. So, again, jw(X(w)) is zero outside of the bandwidth of $X(w)$. So this is correct

Second option: i think it's incorrect, because scaling time domain also scales w domain, which changes badnwidth.

First option: This leads to convolution in the w domain. I think convolution does not preserve bandwidth. Because, say, for some k within the bandwidth, the term $X(k)X(w-k)$ goes outisde the bandwidth.

Is my work right?
Your understanding is right. 

With the usual Fourier transform pair notation :

$$x(t) \longleftrightarrow X(\omega)$$

and assuming bandwidth of $x(t)$ is $W$, then the four cases will be 


1-) $x^2(t) \longleftrightarrow X(\omega) \star X(\omega) \implies Bandwidth = 2W$
2-) $x(2t) \longleftrightarrow \frac{1}{2}X(\omega/2) ~~~~~ ~\implies Bandwidth = 2W$
3-) $ x'(t) \longleftrightarrow j\omega X(\omega) ~~~~~ ~~~~~\implies Bandwidth = W$
4-) $ x(t)\star x(t) \longleftrightarrow X^2(\omega) ~~~\implies Bandwidth = W$


You can see that, cases 3 and 4 has the same bandwidth as $x(t)$ and therefore their Nyquist sampling frequencies will also be the same.