I would live to calculate the primitive roots modulo 26 and modulo 25.
My approach: 26 is not a prime number. But 26=2*13 are Prime numbers. So I calculated the primitive roots of them: Result for 2: 1 Result for 13: 2,6,7,11
So what are the primitive roots of 26? For 25 I know that 25=5*5. But I dont know how it helps.
If ord$\displaystyle_pa=d$ and ord$_2a=1$ if $a$ is odd and $p$ is an odd prime
we can prove that ord$_{2p}a=$lcm$(d,1)=d$
So, if ord$_pa=\phi(p)=p-1$ or ord$_{2p}(a+r\cdot p)=p-1=\phi(2p)$ if $a+r\cdot p$ is odd , where $0\le r<p$
More specifically, if $a$ is a primitive root $\pmod p$
Case $\#1:$ if $a$ is odd and $a$ will be a primitive root $\pmod {2p}$
Case $\#2:$ Else i.e., if $a$ is even, $a+p$ will be a primitive root $\pmod {2p}$