A characterization of 'orthogonal' matrices

Since every nonzero vector is an eigenvector of some symmetric matrix with distinct eigenvalues, and commuting diagonalisable matrices are simultaneously diagonalisable, that $UU^t$ commutes with every symmetric matrix $A$ implies that every nonzero vector is an eigenvector of $UU^t$.

As you said, $UU^t$ commutes with every symmetric matrix $A$. In particular, for every nonzero vector $v$, we have $(UU^tv)v^{\,t}=vv^tUU^t=v(UU^tv)^t$. Therefore, either $UU^tv=0$ or $UU^tv$ is a nonzero scalar multiple of $v$. In either case, $v$ is an eigenvector of $UU^t$.

So, the problem statement boils down to a more fundamental and well known one: if a matrix (not necessarily symmetric) has the property that every nonzero vector is its eigenvector, it must be a scalar multiple of the identity matrix.

A diagonalizable matrix is symmetric iff its eigenspaces are orthogonal. So conjugation by $U$ preserves symmetric matrices iff $U$ preserves orthogonality of arbitrary subspaces of $\mathbb{R}^n$. It is clear that this is equivalent to $U$ preserving orthogonality of individual vectors in $\mathbb{R}^n$.

So, all that remains to be verified is that if $U$ preserves orthogonality of vectors, then it is a multiple of an orthogonal matrix. Now note that $$\langle v+w,v-w\rangle=\langle v,v\rangle-\langle w,w\rangle$$ so two vectors $v$ and $w$ have the same norm iff $v+w$ and $v-w$ are orthogonal. So $U$ preserves when pairs of vectors have the same norm, which implies $U$ multiplies all norms by some constant $\lambda$. Then $\lambda^{-1}U$ preserves norms and hence is orthogonal, so $U$ is a constant multiple of an orthogonal matrix.

[Note that by a similar argument, a complex matrix which preserves Hermitian matrices is a multiple of a unitary matrix. The last part of the argument does not quite work if we are dealing with a complex inner product instead of a real inner product, since the inner product identity used is not true. However, the identity still is true assuming $v$ and $w$ are orthogonal, so that $U$ preserves orthogonal pairs of vectors having the same norm. You can then use this to show $U$ multiplies all norms by a constant by breaking vectors into orthogonal pieces.]