I am confused I have noticed. Perhaps you can clear up my confusion, which I can't pinpoint the origin of.
Let $M$ be a monoid and $a\in M$ such that $a^3=1$, show that the mapping $x\mapsto ax^2a^2$ is an automorphism of $M$
So I know an automorphism to be a self-isomorphism(here $f:M\to M$), that seems alright. I know an isomorphism to be a bijective homomorphism, fair enough. So let's find if we have a homomorphism from above.
A homomorphic map must satisfy $f(xy)=f(x)f(y)$:
$$f(x)f(y)=ax^2a^2ay^2a^2=ax^2a^3y^2a^2=ax^2y^2a^2=f(xy)$$ Happy days, we have the homomorphic property, and we are mapping from $f:M\to M$ so it is an endomorphism(so far).
What about bijection though to upgrade this endomorphism to an automorphism(so to speak), and we must satisfy $f(1_M)=1_M$ for an isomorphism.
Here is where my confusion comes in, $1_M$ I believe is the identity element of $M$, okay, let's find the identity element, we just take the binary operation $x\cdot 1_M = 1_M \cdot x = x$. Now I realise I don't know what my binary operation is. My function takes a single element?
Perhaps I need not show $f(1_M)=1_M$ or perhaps I truly am just confused, any ideas?
Cohn - Classic Algebra - Chapter 3.1 - Monoids - Exercise 11$^*$ - Page 43
$*$I had a mistake in transcribing the exercise, and hence no fault of Cohn - $x \mapsto axa^2$ - Apologies!
Even if $M$ is a group (not just a monoid), and even if $M$ is commutative (not specified), the conclusion is false. Let $M = \mathbb{C}^*$, the group of nonzero complex numbers under multiplication, and let $a = e^{2\pi i / 3}$, so that $a^3 = e^{2\pi i} = 1$. Now define $f: M \to M$ by $f(x) = ax^2 a^2$.
Then $f$ is indeed a (group) homomorphism, but it is not one-to-one: $$ f(-1) = a (-1)^2 a^2 = a(1) a^2 = a^3 = 1, $$ and $$ f(1) = a(1)^2 a^2 = a^3 = 1. $$
Actually for this example we could restrict to just the circle group, or indeed even the cyclic group of order $6$.