Let $N$ be an even integer. Consider the affine cipher on the space of plaintext messages $\mathcal M=\mathbb Z/N\mathbb Z$ with encryption function $e(m)=am$ where $a\;\epsilon\; \mathbb Z/N\mathbb Z$ and $a\neq1\pmod{N}$. Assume $a$ is invertible modulo $N$.
I know that $0$ is going to be a fixed message as $e(0)\equiv a(0)\equiv 0 \pmod{N}$. Also I know that the other fixed message is going to be of the form $$ e(m_1)\equiv am_1\equiv m_1\pmod{N}$$ $$ m_1\equiv a^{-1}m_1 \pmod{N}$$ but I am not sure how to find this without assuming it exists first. I tried to solve the same equation by: $$ am_1-m_1\equiv m_1(a-1)\equiv 0 \pmod{N}$$ but this is also using the same assumption, right?
Write $N=2M$. Since $a$ is invertible mod $N$, you have $\gcd(a,N)=1$. Hence, $a$ cannot be even since $N$ is even. Thus $a$ is odd and so $a=1+2k$ for some $k$. Then $$ e(M)=aM=(1+2k)M=M + 2kM = M + kN = M \bmod N. $$ But clearly $M$ isn't the class of $0$ so this is your other fixed point.