When is a holomorphic submersion with isomorphic fibers locally trivial?

If you drop the properness hypothesis, there are easy counterexamples. For instance, begin with A^1 with coordinate t, and with A^2 with coordinates x and y. Consider the projection p_1: A^1 x A^2 --> A^1, (t,(x,y)) --> t. Now remove the closed subset C = Z(x(x-t),y) as well as the disjoint closed set which is the singleton {(0,(0,1))}. Let X be the open complement. Then p_1:X --> A^1 is flat, all geometric fibers are isomorphic -- each being the complement of two points in A^2, yet the morphism is not locally trivial.

Regarding Sandor's suggestion, something like this is written up in my joint paper with Johan de Jong on "almost properness" of GIT stacks.

Here is a variant of Jason's example with a proof that it is not even topologically locally trivial. Let $T$ be a (complex) manifold that admits a morphism $\phi$ onto $\mathbb P^1=\mathbb P^1_{\mathbb C}$ (or $S^2$ if you prefer) and there exists a point $a\in T$ with $b=\phi(a)\in \mathbb P^1$ such that $\{a\}=\phi^{-1}(b)$. This is satisfied for example if $\phi={\rm id}_{\mathbb P^1}$. Let $\Gamma\subset T\times \mathbb P^1$ be the graph of $\phi$ and let and $c\in \mathbb P^1, c\neq b$.

Now let $X=T\times \mathbb P^1\setminus \bigg( \Gamma\cup \big(T\times \{b\}\big)\cup \{(a,c)\}\bigg)$ with the natural projection $\pi:X\to T$. Then every fiber of $\pi$ is isomorphic to $\mathbb P^1\setminus \{0,\infty\}\simeq \mathbb C^*\sim S^2\setminus \{P,Q\}$ (for two points $P,Q\in S^2$).

Claim $\pi$ is not topologically locally trivial near $a\in T$.

Proof Suppose $a$ has a neighbourhood $U\subseteq T$ such that $Y:=\pi^{-1}U\simeq U\times S^2\setminus \{P,Q\}$. Then there exists a projection $p:Y\to S^2\setminus \{P,Q\}$. Consider a circle in $S^2\setminus \{P,Q\}$ that's non-trivial in $H_1(S^2\setminus \{P,Q\}, \mathbb Z)$. Since $p$ is an isomorphism between $\pi^{-1}(a)$ and $S^2\setminus \{P,Q\}$, the same circle lives in $\pi^{-1}(a)$ as well. Then the homology class of the circle can be represented by a "small" circle around the point $(a,c)$ (this point is not in $X$!). Next take a "small" ball inside $T\times \mathbb P^1$ with center at $(a,c)$ that contains the previous "small" circle. By the construction of $X$, the intersection of this ball with $X$ is the entire ball except its center $(a,c)$. Therefore the homology class of that "small" circle in $X$ is trivial. However, this is a contradiction, because it was chosen in a way that its image via $p_*$ would be a nontrivial homology class. $\qquad {\rm Q.E.D.}$

Remark I suppose a similar proof works to show that Jason's example is also not locally trivial.

Taking your question to the realm of schemes I think that assuming something like that ${\rm Aut} F$ has a natural scheme structure gives you something that could be considered the algebraic equivalent of this statement.

Here is the argument. You can decide what needs to be assumed in addition to the above to get your goal.

Let $\pi:X\to S$ be a smooth morphism and $F$ a (smooth) variety such that $F\simeq X_s$ for all $s\in S$. Consider the relative ${\rm Isom}$ scheme $$ {\rm Isom}_S(X,F\times S)\to S. $$ This is a problem point as it might not exist. Or rather, the ${\mathscr Isom}$ functor is not necessarily representable. See this answer for a sketch of why this functor is representable for projective families and Torsten Ekedahl's answer to the same question for an easy example when it is not.

Anyway, if $I:={\rm Isom}_S(X,F\times S)$ exists, then consider the base change of your family to $I$, $$ X_I=X\times_S I\to I, $$ and consider the relative ${\rm Isom}$ scheme for the new family: $$ {\rm Isom}_I(X_I,F\times I) \to I. $$ Since the fibers of $X$ are isomorphic to $F$, the fibers of this scheme are isomorphic to ${\rm Aut} F$.

From the definition of the ${\mathscr Isom}$ functor it is clear that $$ {\rm Isom}_I(X_I,F\times I)\simeq {\rm Isom}_S(X,F\times S)\times_SI = I\times_S I, $$ so it admits a natural section over $I$. In other words, $X_I\simeq F\times I$.

So this proof seems to show that if ${\rm Isom}_S(X,F\times S)$ exists, then there is a base-change that trivializes $\pi:X\to S$. As a next step I would try to take multisections of ${\rm Isom}_S(X,F\times S)\to S$ to get a finite cover. You would probably want for each $s\in S$ a multisection that is unramified over $s$ to get an étale local trivialization. If you are happy with a local trivialization in the Euclidean topology (assuming you're working over $\mathbb C$) then this should do it. For the issue of local trivialization in the Zariski topology, see below.

Finally, this proof definitely shows that if $F$ is projective with a finite automorphism group, then $\pi:X\to S$ is étale locally trivial: the projectivity of $F$ implies the existence of ${\rm Isom}_S(X,F\times S)$ and the finiteness of ${\rm Aut} F$ implies that ${\rm Isom}_S(X,F\times S)\to S$ is a finite étale morphism.

Addendum

Regarding the discussion of families that are analytically but not algebraically locally trivial, an important difference is whether they have a section or not.

Assume that $\dim S=1$. Then if $\pi:X\to S$ is a smooth projective family without a section, then it cannot be algebraically locally trivial, since the closure of a section of the trivial part over a Zariski open set would give a section of the family. So, this way it is easy to find tons of examples.

On the other hand, a (quasi-)projective family over a curve will always have multisections, so it will have a section after a base change. So, for algebraically locally trivializing an isotrivial family the best hope is to do it after a finite base change. The above proof shows that if the appropriate $\mathscr Isom$ functor is representable by a quasi-projective scheme then this is doable. In particular, if $\pi:X\to S$ is projective, it can be trivialized with a finite base change (take a multisection of ${\rm Isom}_S(X,F\times S)\to S$).