I am dealing with Riemannian immersions and I am stuck on the following: Given a totally geodesic immerserd surface $S$ on a compact riemannian manifold $M$ with metric $g$, is there another metric $\tilde g \neq g$ on $M$ such that $S$ is isometric immersed (not necessarely totally geodesically) on $(M,\tilde g)$? In this case, what can I say about the second fundamental form of this second immersion? Can I conclude with assumptions that the induced connection on $S$ is the same on both immersions?

3$\begingroup$ If the connection on $M$ is the LeviCivita connection, then the induced connection of an immersion is the LeviCivita connection for the induced metric. Thus in both cases, since the immersion is isometric, the induced connection is the same. $\endgroup$ – Paul Bryan Oct 21 '17 at 1:44
Suppose that $S$ is a pair of disjoint Euclidean 2spheres, say of radius 1 and radius 2. Isometrically immerse to the 3sphere $M$ as totally geodesic spheres of radius 1, intersecting along a totally geodesic curve $C$ lying on each of the 2spheres. The preimage of $C$ in $S$ is a pair of circles of different lengths. No perturbation of the metric on $M$ can give that curve $C$ those two different lengths. So $S$ is not isometrically immersed for any metric on $M$.
Suppose that you do have two immersions, one totally geodesic, the other isometric, $S \to M$. If they agree at one point, and have the same LeviCivita connection induced, they are both isometric immersions. So we can easily invent a counterexample. Take $S$ the plane with standard metric, immersed into 3dimensional Euclidean space by an immersion which is isometric only at the origin, and which puts $S$ into the horizontal 2plane, so totally geodesic. Then the LeviCivita connection of the pullback metric on $S$ cannot be the LeviCivita connection of the original metric on $S$.