Given $a,b \in \mathbb{N}$, is it true that $\phi(\gcd(a,b))=\gcd(\phi(a),\phi(b))$?

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Let $a,b \in \mathbb{N}$ and $d=\gcd(a,b)$. Then we have that $a/d$ and $b/d$ are coprime. I am trying to see if $\phi(a)/\phi(d)$ and $\phi(b)/\phi(d)$ are also coprime ($\phi$ is the Euler totient function).

My thoughts: if I can see that $\phi(d)=\gcd(\phi(a),\phi(b))$ then coprimality follows from the first observation. Since that $x \mid y \Rightarrow \phi(x) \mid \phi(y)$ we have that $\phi(d)$ divides both $\phi(a)$ and $\phi(b)$, so $\phi(d)\le\gcd(\phi(a),\phi(b))$.

Is it possible to conclude? What can I do?

Edit: Is not possible: we have that $$\phi(\gcd(3,4))=\phi(1)=1<\gcd(\phi(3),\phi(4))=\gcd(2,2)=2$$ In general, we have that $\gcd(\phi(n),\phi(n+1))\ge2$ for every $n\ge 3$ since $\phi$ is even for entries greater than $2$, but $\gcd(n,n+1)=1$