Suppose $F\to M\stackrel{\pi}{\to} B$ is a Riemannian submersion with totally geodesic fibers, all manifolds compact. In general, unless $M=B\times F$ is a Riemannian product, the isometry groups of these manifolds is not related in any obvious way. In this direction:
Question 1. Give sufficient conditions so that the isometry group $Iso(B)$ of the base also acts by isometries on the total space $M$.
Question 2. Same for the isometry group $Iso(F)$ of the fiber.
A few remarks:
(i) In general, an isometry of the base (or of the fiber) does not extend to an isometry of the total space. Consider e.g. a Klein bottle $K$ as a $S^1$-bundle and note that any rotation of a fiber $S^1$ cannot extend to a continuous map of $K$.
(iii) By a result of Hermann (see e.g. Besse's book on Einstein mflds), since the fibers are assumeed totally geodesic, the holonomy group is a subgroup of the isometry group of the fiber.
[edit: Please disregard the following remark; see V Kapovitch's answer below.]
(ii) A possible answer for (Q1), ignoring the compactness requirement, is when $\sec_B\leq 0$ and $\pi_1(B)=1$. Note this implies $B$ diffeomorphic to the Euclidean space, hence noncompact. With these assumptions it is not hard to see that an isometry $f:B\to B$ can be lifted to an isometry $\tilde f:M\to M$ by letting $\tilde f(x)$ be the endpoint of the lift of the unique minimal geodesic from $p=\pi(x)\in B$ to $f(p)\in B$. Non-positive curvature is used to guarantee uniqueness of the minimal geodesic from $p$ to $f(p)$ and simply-connectedness simply- connectedness is used to guarantee continuity of the lifted map.
(iii) By a result of Hermann ([incorrect -- see e.g. Besse's book on Einstein mflds), since the fibers are assumeed totally geodesic, the holonomy group is a subgroup of the isometry group of Kapovitch's answer below -- I left the fiber.text only so readers can see this previous mistake.]
