Assumption: if a set $X\times Y$, exists such that $\forall\textbf{u},\textbf{v}\in X\times Y. \left(u_1=v_1\implies u_2=v_2\right)$, then a function $f$ exists such that $f:X\to Y$.
The set of points $\Phi_\mathbb{N}=\left\{(n,\phi(n))\mid n\in\mathbb{N}\right\}$ describing the graph/image of the totient function is a subset of the reals.
By definition, $\forall \textbf{x},\textbf{y}\in\mathbb{R}^2$, where $x_1< y_1.\exists \textbf{z}\in\mathbb{R}^2:x_1<z_1<y_1$. Thus, there is a continuous set of points between any two $\textbf{a},\textbf{b}\in\Phi_\mathbb{N}$. It follows, that there is a a set $\Phi_\mathbb{R}\supset\Phi_\mathbb{N}$ which is the graph of a continuous function $\phi:\mathbb{R}\to\mathbb{R}$, ($n\in\mathbb{N}\implies \phi(n)\in\mathbb{N}$).
Thus, it is at least possible to extend the totient function to the reals without breaking the original definition. Is there a canonical way to do this? If not, what is the most appropriate way to extend the totient function?
Note: An extension to the complex numbers is perfectly acceptable too, I just assumed it would be best to "keep it real".
Note: tagged analysis because that's where I'm going with this.
Edit: This might help... probably not, though
For coprime $a$ and $n$ $$a^{\phi(n)}\equiv1\mod{n}\qquad\text{(Euler's Theorem)}$$ For $x,y\in\mathbb{R}$ $$x\ \text{mod}\ y^{*}=\frac{|x|}{2\pi}\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kix}{y}\right)\right]$$ So $$a^{\phi(n)}\ \text{mod}\ n=\frac{|a^{\phi(n)}|}{2\pi}\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kia^{\phi(n)}}{n}\right)\right]=1\implies$$ $$|a^{\phi(n)}|=2\pi\left(\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kia^{\phi(n)}}{n}\right)\right]\right)^{-1}=$$ $$-4\pi i\left(\ln\left(1-\exp\left(\frac{2\pi ia^{\phi(n)}}{n}\right)\right)-\ln\left(\exp\left(-\frac{2\pi ia^{\phi(n)}}{n}\right)\left(-1+\exp\left(\frac{2\pi ia^{\phi(n)}}{n}\right)\right)\right)\right)^{-1}$$
Assuming $a^{\phi(n)}$ is positive, let $a^\phi(n)=u$ (because I'm running out of space)
$$u=-4\pi i\left(\ln\left(1-\exp\left(\frac{2\pi iu}{n}\right)\right)-\ln\left(\exp\left(-\frac{2\pi iu}{n}\right)\left(-1+\exp\left(\frac{2\pi iu}{n}\right)\right)\right)\right)^{-1}$$
Solve for for $u$ (not sure how), then substitute $a^{\phi(n)}$ back in and take the base-$a$ logarithm to get $\phi(n)$
Assuming all the algebra is correct, this might work...
$^*$ If I understand correctly, we can translate $x\equiv z\mod{y}$ to $x\ \text{mod}\ y=z$, using $\text{mod}$ as a binary operation.
Let's first try to define $\phi$ on $\Bbb Q$; presumably, any canonical definition on $\Bbb R$ would have a restriction to $\Bbb Q$ that would meet with our focus-on-$\Bbb Q$ approval. It will be convenient to define $f(n):=n^{-1}\phi(n)$.
For $n\in\Bbb N$ we have $f(n)=\prod_{p\in\Bbb P\land p|n}(1-p^{-1})$, and it is natural to define $\phi(0)=0$ (since there are no integers from $1$ to $0$ exclusive to be coprime to $0$) and, if we want to extend to negative integers, presumably $f(-n)=f(n)$ since the same primes divide $-n$. If for coprime $m,\,n\in\Bbb Z$ with $n>0$ we demand $f(m/n)=f(m)f(n)$ [sic], so as to include $1-1/p$ factors for every prime appearing in $m/n$'s prime factorisation, we have a canonical extension to $\Bbb Q$. In particular, this definition implies $\phi(p^k)=p^k(1-p^{-1})$ for $p\in\Bbb P,\,k\in\Bbb Z\backslash\{0\}$; unsurprisingly, as $p^k\to 0$ we also have $\phi\to 0$.
However, I don't think we can extend this to $\Bbb R$. The only obvious definition is $\phi(x):=\lim_{n\to\infty}\phi(x_n)$ for rationals $x_n$ satisfying $\lim_{n\to\infty}x_n=x$. But I would be very surprised if, for all irrational $x$, this definition is independent of the sequence chosen. In fact, $\phi$ as defined above isn't even continuous at $0$. For example, take your favourite distinct primes $p,\,q$, then define $x_n:=p^n/q^{a_n}$, with $a_n$ just large enough that $x_n<1/n$. Then $\phi(x_n)=(1-1/p)(1-1/q)$ for all $n$.