According to this proof, given an expression $$x^e\pmod n$$ the modulus of the exponent $e$ is $\phi(n)$.
From Euler's Theorem, I know that $$x^{\phi(n)}\equiv 1\pmod n$$ holds true iff $x$ is coprime with $n$.
Is it is always the case that the exponent in a$\mod n$ expression is $\phi(n)$, or does this only apply when $x$ and $n$ are coprime ?
If it only applies when $x$ and $n$ are coprime then does that mean that $e$ can only be reduced when $x$ and $n$ are coprime?
On
When $(x, n) = 1$, we have two different concepts:
Note that $r \mid \phi(n)$ is always true, but this does not mean $r = \phi(n)$. If you have some $e$ such that $x^e \equiv 1 \pmod{n}$, then you can say $r \mid e$ and also $(e, \phi(n)) \geq r$, but that's about it.
Perhaps you would also be interested in the fact that $x^e \equiv x^{e \pmod{r}} \equiv x^{e \pmod{\phi(n)}} \pmod{n}$, which is a consequence of the facts above. Note that none of the points above are valid if $(x, n) \neq 1$; in such a case, $x^e \equiv 1 \pmod{n}$ has no solutions.
The question has nothing to do with powers of $x$ being congruent to 1 modulo $n$. Rather, the question is: is it true that $x^a \equiv x^b \mod n$ whenever $a \equiv b \mod \phi(n)$, for all $x$. When $x$ and $n$ are not relatively prime, this is WEAKER than anything similar to Euler's theorem.
The answer is NO, this is not generally true. Take $x=2$ and $n=4$; then $x^b \equiv 0$ for all $b \ge 2$, so NO power of $x$ greater than or equal to $2$ is congruent to $x \mod 4$. So $x^{\phi(4) + 1} \not\equiv x \mod 4$$ $. You don't even need to know what $\phi(4)$ is; it's $2$, but that is irrelevant. You can use the language about "exponent is modulo $\phi(n)$" ONLY when $x$ and $n$ are relatively prime.