Just a non-thought-out thought:

For any finite simple group $S$, consider the set of maps from $\pi$ to $S$ up to conjugacy. Now, $Out(\hat{\pi})$ acts on this set. Now consider the composition of this action with the permutation character; you get a character $f(S)$ of $Out(\hat{\pi})$ valued in $\pm 1$. [If $S$ is $\mathbf{Z}/p\mathbf{Z}$, I think but didn't check that the corresponding character is the "determinant composed with the quadratic residue symbol mod $p$," if that makes sense.]

I have no idea as to the image of the map $F = \prod_{S} f(S)$, but it seems plausible to me that it is uncountable. On the other hand, since $Out(\pi)$ is finitely generated, the restriction of $F$ to it must have finite image. (Slight clarification: restrict the product over $S$ to nonabelian finite simple groups, since the abelian ones provide no new information.)

Just a non-thought-out thought:

For any finite simple group $S$, consider the set of maps from $\pi$ to $S$ up to conjugacy. Now, $\mathrm{Out}(\widehat{\pi})$ acts on this set. Now consider the composition of this action with the permutation character; you get a character $f(S)$ of $\mathrm{Out}(\widehat{\pi})$ valued in $\pm 1$. [If $S$ is $\mathbb{Z}/p\mathbb{Z}$, I think but didn't check that the corresponding character is the "determinant composed with the quadratic residue symbol mod $p$," if that makes sense.]

I have no idea as to the image of the map $F = \prod_{S} f(S)$, but it seems plausible to me that it is uncountable. On the other hand, since $\mathrm{Out}(\pi)$ is finitely generated, the restriction of $F$ to it must have finite image. (Slight clarification: restrict the product over $S$ to nonabelian finite simple groups, since the abelian ones provide no new information.)

Just a non-thought-out thought:

For any finite simple group $S$, consider the set of maps from $\pi$ to $S$ up to conjugacy. Now, $Out(\hat{\pi})$ acts on this set. Now consider the composition of this action with the permutation character; you get a character $f(S)$ of $Out(\hat{\pi})$ valued in $\pm 1$. [If $S$ is $\mathbf{Z}/p\mathbf{Z}$, I think but didn't check that the corresponding character is the "determinant composed with the quadratic residue symbol mod $p$," if that makes sense.]

I have no idea as to the image of the map $F = \prod_{S} f(S)$, but it seems plausible to me that it is uncountable. On the other hand, since $Out(\pi)$ is finitely generated, the restriction of $F$ to it must have finite image.

Just a non-thought-out thought:

For any finite simple group $S$, consider the set of maps from $\pi$ to $S$ up to conjugacy. Now, $Out(\hat{\pi})$ acts on this set. Now consider the composition of this action with the permutation character; you get a character $f(S)$ of $Out(\hat{\pi})$ valued in $\pm 1$. [If $S$ is $\mathbf{Z}/p\mathbf{Z}$, I think but didn't check that the corresponding character is the "determinant composed with the quadratic residue symbol mod $p$," if that makes sense.]

I have no idea as to the image of the map $F = \prod_{S} f(S)$, but it seems plausible to me that it is uncountable. On the other hand, since $Out(\pi)$ is finitely generated, the restriction of $F$ to it must have finite image. (Slight clarification: restrict the product over $S$ to nonabelian finite simple groups, since the abelian ones provide no new information.)