Does exist any integer $n\geq 1$ satisfying $\varphi(12n)=\varphi(6n+5)$, where $\varphi(m)$ denotes the Euler's totient function?

Question

We consider the Euler's totient function $\varphi(n)$ for integers $n\geq 1$. I was playing with the condition $$\varphi(12n)=\varphi(6n+a)\tag{1},$$ for different and simple (integer) values of $a$, when I wondered if one can to find some solution of the equation $$\varphi(12n)=\varphi(6n+5).\tag{2}$$

Question. I don't know why for some integers like $a=1,2,3$ or $a=4$ one can find solutions of $(1)$, but for different values likes than our equation $(2)$ it is difficult to find solutions. Do you have an idea about it? Specially I am asking if we can find some integer $n\geq 1$ of $$\varphi(12n)=\varphi(6n+5).$$ Many thanks.

Computational fact. There aren't solutions of $(2)$ in the interval $1\leq n\leq 10^7.\square$

I think that this problem $(2)$ isn't in the literature, any case if you know it, answer this question as a reference request and I try to find and read the literature to know if there exist some solution of $(2)$

Answer 1

There is at least one solution as this PARI/GP-program shows :

? for(n=1,10^8,if(eulerphi(12*n)==eulerphi(6*n+5),print(n)))
80538624
?

Upto $10^8$, there is apparently only one solution.