Suppose you have three positive integers $a, b, c$ that form a Pythagorean triple:
\begin{equation}
a^2 + b^2 = c^2. \tag{1}\label{1}
\end{equation}
Additionally, suppose that when you apply Euler's totient function to each term, the equation still holds:
$$
\phi(a^2) + \phi(b^2) = \phi(c^2). \tag{2}\label{2}
$$
One way this can happen is if $a^2, b^2, c^2$ have the same primes in their prime factorization. (For example, starting from the Pythagorean triple $3,4,5$, we could multiply all three terms by $30$ to get $90, 120, 150$. If we do, then we have $90^2 + 120^2 = 150^2$ and $\phi(90^2) + \phi(120^2) = \phi(150^2)$.) In that case, because all three terms are squares, they all contain these prime factors at least twice, and so we must have
$$
\phi(\phi(a^2)) + \phi(\phi(b^2)) = \phi(\phi(c^2)). \tag{3}\label{3}
$$
My question is: are there any "atypical" solutions to the two equations $\eqref{1}$ and $\eqref{2}$ for which $\eqref{3}$ does *not* hold? Or at least where $\eqref{1}$ and $\eqref{2}$ hold, but the prime factorizations of $a,b,c$ do not consist of the same primes, even if $\eqref{3}$ happens to hold for a different reason?

In the comments, Peter and Gerry Myerson have checked small cases (all triples for $1 \le a \le b \le 10^5$ and primitive triples generated by $(m,n)$ for $1 \le n \le m \le 2000$) without finding any atypical solutions.

Here is an in-depth explanation for why typical solutions like $(90,120,150)$ work. By a typical solution, I mean a solution where $a,b,c$ have the same primes in their prime factorization. Such a triple satisfies $\eqref{2}$ and $\eqref{3}$ whenever it satisfies $\eqref{1}$, as shown below.

Let $\operatorname{rad}(x)$ denote the radical of $x$: the product of all distinct prime factors of $x$. To get a typical solution, we start with any Pythagorean triple, then scale $(a,b,c)$ so that $\operatorname{rad}(a) = \operatorname{rad}(b) = \operatorname{rad}(c) = r$.

It is a general totient function identity that whenever $\operatorname{rad}(x) = r$, $\phi(x) = \frac{\phi(r)}{r} \cdot x$. In other words, $\phi(x) = x \prod\limits_{p \mid x} \frac{p-1}{p}$ where the product is over all primes $p$ that divide $x$.

In the case above, we have
$$
\phi(a^2) + \phi(b^2) = \frac{\phi(r)}{r} \cdot a^2 + \frac{\phi(r)}{r} \cdot b^2 = \frac{\phi(r)}{r} \cdot c^2 = \phi(c^2),
$$
and $\eqref{2}$ holds.
Moreover, since $r \mid a,b,c$, we have $r^2 \mid a^2,b^2,c^2$, so when we multiply by $\frac{\phi(r)}{r}$, we have $r \phi(r) \mid \phi(a^2), \phi(b^2), \phi(c^2)$. Therefore all prime factors of $r \phi(r)$ divide each of $\phi(a^2)$, $\phi(b^2)$, and $\phi(c^2)$. These are *all* their prime factors, since $r$ contained all the prime factors of $a^2, b^2,c^2$ and since then the only new prime factors introduced came from multiplying by $\phi(r)$.

As a result, $\phi(a^2), \phi(b^2), \phi(c^2)$ still have the same set of prime factors: $\operatorname{rad}(\phi(a^2)) = \operatorname{rad}(r \phi(r)) = s$, and similarly $\operatorname{rad}(\phi(b^2)) = \operatorname{rad}(\phi(c^2)) = s$. So $\eqref{3}$ holds, because $$ \phi(\phi(a^2)) + \phi(\phi(b^2)) = \frac{\phi(s)}{s} \cdot \phi(a^2) + \frac{\phi(s)}{s} \cdot \phi(b^2) = \frac{\phi(s)}{s} \cdot \phi(c^2) = \phi(\phi(c^2)). $$