It is not possible in general for $\,n \ge 4\,.$ By Vieta's relations:

$$ \hat{p}_1(t)=t^n + r_1 a_{n-1}t^{n-1}+r_1^2a_{n-2}t^{n-2}+\ldots +r_1^{n-1}a_1t + r_1^na_0 \\ \hat{p}_2(t)=t^n + r_2 b_{n-1}t^{n-1}+r_2^2b_{n-2}t^{n-2}+\ldots +r_2^{n-1}b_1t + r_2^nb_0 \\ $$

Therefore $\,r_1,r_2\,$ can be chosen such that at most two of the coefficients match between $\,\hat{p_1}(t)\,$ and $\,\hat{p_2}(t)\,$, but in general not more than that.  "At most" because in case of even powers of $\,r_{1,2}\,$ the respective coefficients must also have the same signs, then there are aditional restrictions if some of the coefficients are zero.

It is not possible in general for $\,n \gt 2\,.$ By Vieta's relations:

$$ \hat{p}_1(t)=t^n + r_1 a_{n-1}t^{n-1}+r_1^2a_{n-2}t^{n-2}+\ldots +r_1^{n-1}a_1t + r_1^na_0 \\ \hat{p}_2(t)=t^n + r_2 b_{n-1}t^{n-1}+r_2^2b_{n-2}t^{n-2}+\ldots +r_2^{n-1}b_1t + r_2^nb_0 \\ $$

For two coefficients to end up being equal the condition is $\,r_1^ia_{n-i}=r_2^ib_{n-i}\,$. Another pair of coefficients can become equal iff $\,r_1^ja_{n-j}=r_2^jb_{n-j}\,$. But (assuming non-zero coefficients) the two equations are consistent only iff $\,\left(\dfrac{b_{n-i}}{a_{n-i}}\right)^j=\left(\dfrac{b_{n-j}}{a_{n-j}}\right)^i=\left(\dfrac{r_1}{r_2}\right)^{ij}\,$. Therefore, it is not possible to match more than one pair of coefficients with this construct in general, so $\,n \le 3\,$.

The case $\,n=3\,$ is excluded because the coefficients of the linear terms can be made to match $\,r_1^2a_1=r_2^2b_1\,$ only if $\,a_1\,$ and $\,b_1\,$ have the same sign.

It is not possible in general for $\,n \ge 4\,.$ By Vieta's relations:

$$ \hat{p}_1(t)=t^n + r_1 a_{n-1}t^{n-1}+r_1^2a_{n-2}t^{n-2}+\ldots +r_1^{n-1}a_1t + r_1^na_0 \\ \hat{p}_2(t)=t^n + r_2 b_{n-1}t^{n-1}+r_2^2b_{n-2}t^{n-2}+\ldots +r_2^{n-1}b_1t + r_2^nb_0 \\ $$

Therefore $\,r_1,r_2\,$ can be chosen such that at most two of the coefficients match between $\,\hat{p_1}(t)\,$ and $\,\hat{p_2}(t)\,$, but in general not more than that.  "At most" because in case of even powers of $\,r_{1,2}\,$ the respective coefficients must also have the same signs, then there are aditional restrictions if some of the coefficients are zero.

It is not possible in general for $\,n \ge 4\,.$ By Vieta's relations:

$$ \hat{p}_1(t)=t^n + r_1 a_{n-1}t^{n-1}+r_1^2a_{n-2}t^{n-2}+\ldots +r_1^{n-1}a_1t + r_1^na_0 \\ \hat{p}_2(t)=t^n + r_2 b_{n-1}t^{n-1}+r_2^2b_{n-2}t^{n-2}+\ldots +r_2^{n-1}b_1t + r_2^nb_0 \\ $$

Therefore $\,r_1,r_2\,$ can be chosen such that at most two of the coefficients match between $\,\hat{p_1}(t)\,$ and $\,\hat{p_2}(t)\,$, but in general not more than that.  "At most" because in case of even powers of $\,r_{1,2}\,$ the respective coefficients must also have the same signs, then there are aditional restrictions if some of the coefficients are zero.