$$P(c \cdot x) = \cos(x) P(x)$$
For $c=2$, $P(x) = \sin(x)/x$ is a solution to this. I don't know if there's a closed-form solution for $c \ne 2$.
Rather than add my own attempt at solution, which is dubious at best, I'll add context. I've been trying to solve a rather odd line integral here, and I ended up with a very nice infinite product solution. Robert Israel, put this solution into the form of recurrence relation. So, in attempt to continue the wild-goose-chase, I'm pursuing a solution for this recurrence relation.
(I'll take any solution, as long as the "special functions" used are not tautological and have a public-access paper written about them)
We know that $c \neq 0$, unless $P(x) = 0$ for all $x$. Also $c\neq 1$, since otherwise we have $P(x)=\cos(x)P(x) \implies \cos(x) =1$ for all $x$.
If it is assumed that $P$ can be expressed as a power series centered at the origin (supported by the result for $c=2$) we find:
$$\sum_{n=0}^\infty a_n c^n x^n = \cos(x) \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!} \sum_{n=0}^\infty a_n x^n.$$
Thus $$a_n c^n = \sum_{i=0}^n a_i b_{n-i}$$ where $$b_{i} = \left\{ \begin{array}{cc} (-1)^{i/2}/(i)! & i \text{ even}\\ 0 & i \text{ odd}\end{array}\right.$$
Writing out each series to compare terms gives:
$$a_0 + a_1cx+a_2c^2x^2+\cdots = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) \left( a_0 + a_1x + a_2x^2 +a_3 x^3 + \cdots \right)$$
Thus $a_0 = a_0$, which provides some degree of freedom of choice for $a_0$. (Note that $A\sin(x)/x$ will satisfy $P(2x)=\cos(x)P(x)$ as well).
Then $a_1 c = a_1$. Either this means that $c=1$ or $a_1=0$. Since we have shown that $c\neq 1$ we have $a_1=0$.
Continuing on this line gives us $$a_2 c^2 = a_2 - \frac{a_0}{2!}$$ which yields $$a_2 = \frac{-a_0}{2!(c^2-1)}.$$
If you continue this process for $a_3, a_4,...$, you will find that $a_{2n+1}=0$ for all $n$, and you will find expressions for each $a_{2n}$ in terms of $a_0$. There might be a nice closed form expression for $a_{2n}$, but I haven't committed to finding it.