A numeric example of suitable isosceles triangle:

\begin{align}
R&=1
,\\
\alpha&\approx 18.5528612^\circ
,\quad
\beta=\gamma \approx 80.7235694^\circ
,\\
a&\approx 0.6363589
,\quad
b=c\approx 1.9738442
,\\
r&\approx 0.2704262
,\\
a+b+b&\approx 4.5840473
,\\
2\pi(R-r)&\approx 4.5840473
.
\end{align}
Edit:
Another example is an obtuse isosceles triangle:

\begin{align}
R&=1
,\\
\alpha&\approx 106.8724079^\circ
,\quad
\beta=\gamma \approx 36.56379605^\circ
,\\
a&\approx 1.9139069
,\quad
b=c\approx 1.1914349
,\\
r&\approx 0.3161467
,\\
a+b+b&\approx 4.2967768
,\\
2\pi(R-r)&\approx 4.2967768
.
\end{align}
Edit:
And these are the only two suitable shapes of isosceles triangles.
This can be shown as follows.
Using known identities for triangles
\begin{align}
\sin\alpha+\sin\beta+\sin\gamma&=\frac{a+b+c}{2R}
,\\
\cos\alpha+\cos\beta+\cos\gamma&=\frac{r}{R}+1
.
\end{align}
For isosceles triangle we have
\begin{align}
\sin2\beta+2\sin\beta&=\frac{a+b+c}{2R}
,\\
-\cos2\beta+2\cos\beta&=\frac{r}{R}+1
,\\
\end{align}
and condition $a+b+c=2\pi(R-r)$
can be expressed in terms of $\beta$ as follows:
\begin{align}
2\sin\beta\cos\beta+2\sin\beta -
\pi(1+2\cos^2\beta-2\cos\beta)=0
,
\end{align}
which can be simplified to
\begin{align}
5\pi t^4-2\pi t^2-8t+\pi&=0
,\qquad t=\tan\tfrac\beta2
.
\end{align}
For $\beta\in(0,\tfrac\pi2)$,
this equation results in two real values
$t_1\approx 0.3303679220$,
$t_2\approx 0.8499172732$
with corresponding values of $\beta$,
shown above.
Edit:
Using that the sides $a,b,c$ of a triangle
with semiperimeter $\rho=\tfrac12(a+b+c)$,
incircle $r$ and circumcircle $R$
are the roots of cubic polynomial
\begin{align}
x^3&-2\rho x^2+(\rho^2+4rR+r^2)x-4\rho rR
\tag{g1}\label{g1}
\end{align}
and scaling the triangle such that $R=1$,
the condition $2\rho=2\pi(R-r)$ converts \eqref{g1} to
\begin{align}
x^3-2\pi\,(1-r)\,x^2+((\pi^2+1)\,r^2+(4-2\pi^2)\,r+\pi^2)\,x-4\pi\,r(1-r)
\tag{g2}\label{g2}
.
\end{align}
Recall that the cubic polynomial $ax^3+bx^2+cx+d$ has discriminant
\begin{align}
\Delta&=b^2c^2-27 a^2 d^2-4(a c^3+b^3 d)+18 abcd
,\\
\text{and in case of \eqref{g2}}\quad
\Delta=
&-(\pi^2+1)^2\,r^4
\\
&+4\,(\pi^4+6\,\pi^2-3)\,r^3
\\
&-2\,(19\,\pi^2+3\,\pi^4+24)\,r^2
\\
&+4\,(\pi^4+3\,\pi^2-16)\, r
\\
&-\pi^2(\pi^2-4)
\tag{g3}\label{g3}
,\\
\Delta&>0\quad \forall\ r\in(0.2704262, 0.3161467)
,
\end{align}
that means, there are infinitely many scalene triangles,
which inradius is in this range,
as well as there also are infinitely many triangles
with $r\in(0,0.2704262)\cup(0.3161467,0.5)$
for which the condition $2\rho=2\pi(R-r)$ does not hold.
For example, let's take $r=0.29$.
Solutions to
\begin{align}
\\
\text{gives }\quad&
a\approx 0.7586625863,\quad b\approx 1.722967032,\quad c\approx 1.979431950
,\\
a+b+c&\approx 4.461061568
,\\
2\pi(R-r)&\approx 4.461061568
.
\end{align}

Also note that the two endpoints of the $r$-interval
correspond to the pair of isosceles triangles shown above.