Functional/Differential Equation In the midst of a calculation too long (and too irrelevant) to describe here, I've been forced to confront the following equation:

$$1-p-f(f(p))-f(p)f'(f(p))=0$$

Here $f$ is a differentiable function from the unit interval to itself, and the equation is supposed to hold identically in $p$.  How do I find all such functions $f$?
$\large\mbox{At least, two solutions.}$

Let's ${\rm f}\left(p\right) \equiv \mu p + \nu$.
$$
1 - p - \mu\left(\mu p + \nu\right) - \nu - \left(\mu p + \nu\right)\mu = 0
$$

$$
\left(1 - \mu\nu - \nu - \nu\mu\right)
+
\left(-1 - \mu^{2} - \mu^{2}\right)p
=
0
$$

$$
\mu = \pm\,{\sqrt{2} \over 2\,}\,{\rm i}\,,
\qquad
\nu
=
{1 \over 2\mu + 1}
=
{1 \over \pm\,\sqrt{2\,}\,{\rm i} + 1}
=
{1 \over 3}\,\left(1 \mp \sqrt{2\,}\,{\rm i}\right)
$$

$$
\color{#ff0000}{\large%
{\rm f}_{-}\left(p\right)
\color{#000000}{\ =\ }
-\,{\sqrt{2\,} \over 2}\,{\rm i}\,p
+
{1 \over 3}\,\left(1 + \sqrt{2\,}\,{\rm i}\right)\,,
\qquad
{\rm f}_{+}\left(p\right)
\color{#000000}{\ =\ }
{\sqrt{2\,} \over 2}\,{\rm i}\,p
+
{1 \over 3}\,\left(1 - \sqrt{2\,}\,{\rm i}\right)}
$$

$\large\mbox{Just trying to guess:}$

Let's $\quad x \equiv {\rm f}\left(p\right)$. Then,
$$
x{\rm f}'\left(x\right) + {\rm f}\left(x\right)
=
1 - p\left(x\right)\,,
\quad
x{\rm f}\left(x\right)
-
a{\rm f}\left(a\right)
=
x - a
-
\int_{a}^{x}p\left(x'\right)\,{\rm d}x'
$$

$$
{\rm f}\left(x\right)
=
{a\left[{\rm f}\left(a\right) - 1\right] \over x}
+
1
-
{1 \over x}\int_{a}^{x}p\left(x'\right)\,{\rm d}x'
$$
$\large\mbox{At least, two solutions.}$

Let's ${\rm f}\left(p\right) \equiv \mu p + \nu$.
$$
1 - p - \mu\left(\mu p + \nu\right) - \nu - \left(\mu p + \nu\right)\mu = 0
$$

$$
\left(1 - \mu\nu - \nu - \nu\mu\right)
+
\left(-1 - \mu^{2} - \mu^{2}\right)p
=
0
$$

$$
\mu = \pm\,{\sqrt{2} \over 2\,}\,{\rm i}\,,
\qquad
\nu
=
{1 \over 2\mu + 1}
=
{1 \over \pm\,\sqrt{2\,}\,{\rm i} + 1}
=
{1 \over 3}\,\left(1 \mp \sqrt{2\,}\,{\rm i}\right)
$$

$$
\color{#ff0000}{\large%
{\rm f}_{-}\left(p\right)
\color{#000000}{\ =\ }
-\,{\sqrt{2\,} \over 2}\,{\rm i}\,p
+
{1 \over 3}\,\left(1 + \sqrt{2\,}\,{\rm i}\right)\,,
\qquad
{\rm f}_{+}\left(p\right)
\color{#000000}{\ =\ }
{\sqrt{2\,} \over 2}\,{\rm i}\,p
+
{1 \over 3}\,\left(1 - \sqrt{2\,}\,{\rm i}\right)}
$$

$\large\mbox{Just trying to guess:}$

Let's $\quad x \equiv {\rm f}\left(p\right)$. Then,
$$
x{\rm f}'\left(x\right) + {\rm f}\left(x\right)
=
1 - p\left(x\right)\,,
\quad
x{\rm f}\left(x\right)
-
a{\rm f}\left(a\right)
=
x - a
-
\int_{a}^{x}p\left(x'\right)\,{\rm d}x'
$$

$$
{\rm f}\left(x\right)
=
{a\left[{\rm f}\left(a\right) - 1\right] \over x}
+
1
-
{1 \over x}\int_{a}^{x}p\left(x'\right)\,{\rm d}x'
$$