Does there exist homomorphism $\phi:\Bbb Z_{24}\to\Bbb Z_{24}$ such that $\phi(14)=15$?

Question

My thoughts:

Since $\Bbb Z_{24}$ is cyclic, generated by $a=1$ $\phi(14)=14 \phi(a)=15 \mod(24) $.

Does that mean that $\phi(a)$ is the solution of the equation $\phi(a)=14^{-1}15 \pmod{24} $?

Thanks.

Answer 1

I'm going to use additive notation, so $\mathbb{Z}_{24} = \mathbb{Z}/24\mathbb{Z}.$

Suppose $\phi(14) = 15$, then $\phi(14\times12) = 12\phi(14) = 12 \times 15 = 12$. At the same time, $14 \times 12$ is a multiple of $24$, hence $\phi(14 \times 12) = 0$. But $12 \neq 0$.