If there are not, then would it be easier to say that 2 objects are identical as ordered fields as opposed to being isomorphic as ordered fields? Or is the word isomorphism used to emphasise the fact that the objects are different as sets?

32$\begingroup$ Yes, it is a Bad Idea to identify a finitedimensional vector space with its dual. $\endgroup$– Andy PutmanMay 12, 2010 at 2:03

$\begingroup$ Is it safe to identify a Hilbert space with its dual? $\endgroup$– teilMay 12, 2010 at 3:21

4$\begingroup$ There is a natural map from a Hilbert space to its dual, but it's antilinear. $\endgroup$– Qiaochu YuanMay 12, 2010 at 3:26

1$\begingroup$ The whole idea of grupoids, stacks and $(\infty, 1)$stacks is to consider higher automorphisms to get a universal object. So automorphisms are important... $\endgroup$– user40276Mar 21, 2015 at 22:24

2$\begingroup$ Riesz's representation theorem is a great thing indeed, but it's the source of 90% of misconceptions physicists have about mathematical matters. :) $\endgroup$– user41593Mar 21, 2015 at 22:27
11 Answers
Inside of the complex numbers there are lots of examples of distinct fields which are isomorphic. For instance, there are three subfields of the form ${\mathbf Q}(\alpha)$ where $\alpha^3 = 2$: take for $\alpha$ any of the three complex cube roots of 2 and you get a different subfield. What are the consequences of treating them as literally equal? You can't make any sense of Galois theory if you do that! Similarly, all $p$Sylow subgroups of a finite group are isomorphic (since conjugate subgroups are isomorphic groups), but it would kind of destroy a lot of the content of the Sylow theorems by trying to say the $p$Sylow subgroups are identical.
More generally, anytime you have isomorphic but unequal objects inside a larger object, it can lead to confusion if not outright incomprehensibility if you try to regard them all as identical. (There was a paper by Chevalley about unit groups in number fields where he made a genuine error by an abuse of the "square root" notation and I think one might be able to express the mistake in the form of an isomorphism being confused with an equality, but I'd have to look at the paper again to be sure about this.)
The word isomorphism does not emphasize that two objects are different; any group or vector space admits an isomorphism with itself using the identity map. The word emphasizes that in a structural way the two objects look like each other even though they are not literally the same. Never say two objects are identical if they are not actually identical. Having said that, I must admit that in mathematics one meets phrases like "since $X$ and $Y$ are isomorphic we can identify $X$ with $Y$" and then $X$ is replaced with $Y$. The usefulness of doing this depends on the application you have in mind. Note, however, that replacing $X$ with $Y$ is not saying that $X$ and $Y$ are the same thing.
This question sounds like it is being asked by someone who hasn't had a lot of experience with isomorphisms and is trying to get a feel for what it means. In a year or two, after seeing more appearances of the concept and its uses, you'll get a better feel for it, but for now do not think the word isomorphism is a synonym for identical.

9$\begingroup$ A very simple example for the danger of identifying isomorphic subobjects being “$ℤ = 2ℤ$, so $ℤ/2ℤ = ℤ/ℤ = 0$.” $\endgroup$– k.stmNov 23, 2014 at 15:43
A slightly more general answer: if the objects in question have no nontrivial automorphisms (i.e. nonidentity isomorphisms from itself to itself) then no danger will come from treating isomorphisms as the identity. (E.g. the real numbers has no automorphisms as a field, and so any two copies of the real numbers can be unambiguously identified as fields.)
But if $X$ and $Y$ are isomorphic but Aut($Y$) (or, equivalently, Aut($X$)) is nontrivial, then the identification of $X$ and $Y$ is not uniquely determined (because you could always postcompose with a nontrivial automorphism of $Y$, or precompose with a nontrivial automorphism of $X$, to obtain a different identification), and hence in this case there is not an unambiguous identification. (A typical example of this is the isomorphism between a finitedim'l vector space and its dual; there is not a uniquely determined such isomorphism, and so one should not identify the two, although they are isomorphic.)
In some contexts, although there is not a uniquely determined isomorphism, there is a canonical choice, e.g. the identification of a finite dimensional space with its double dual. Such isomorphisms are normally described in the language of natural isomorphisms between functors (e.g. the identify functor and the double duality functor are naturally isomorphic as functors from the category of finitedimensional vector spaces  over some given field  to itself.) Often it is safe to identify objects that are identified by a natural isomorphism in some category theoretic framework that is suitable for the problem at hand. But even in these cases, sometimes one has to actually know an explicit description of the natural isomorphism (e.g. because you might need to make a computation involving both objects, which will of course require you to know how they have been identified). In particular, when identifying two objects via some natural isomorphism, it is good form to explicitly describe the natural isomorphism at some point, unless the natural isomorphism in question is utterly conventional (e.g. the double duality isomorphism for a finitedimensional vector space; and even then, one might write something like "we identify $V$ and $V^{\vee\vee}$ via the usual double duality isomorphism").
Added: This answer is very similar to that of Charles Staats, which was posted while I was writing.

9$\begingroup$ Even canonical, natural isomorphisms can lead to problems. For example, $A\times B\cong B\times A$ is not the identity for $A=B$. $\endgroup$– user2035May 12, 2010 at 11:35

1$\begingroup$ It is possible for nonisomorphic rigid structures to become isomorphic by a unique isomorphism in a forcing extension. In this case, your view would have us say that nonidentical objects become identical when enlarging the universe. I think this way of thinking will be very confusing. $\endgroup$ May 12, 2010 at 17:38
If $p$ is a prime, then the multiplicative group modulo $p$ is isomorphic to the additive group modulo $p1$, which is to say it's a cyclic group of order $p1$. It's trivial to find a generator of the additive group, but a fair bit of work to find one for the multiplicative. Related to this, it's easy to solve $ax=c$ in the additive group, using the Euclidean algorithm, but the difficulty of solving the corresponding equation $a^x=c$ in the multiplicative group  the discrete logarithm problem  is the basis for some cryptography systems.
First of all, as other posters have said, you never claim that two objects are identical unless they actually are equal as sets. However, there are times when it is appropriate to "identify" two isomorphic objects. (This is really just an issue of language).
Exactly when such identification is appropriate is a matter of some subtlety. If there are many different isomorphisms between two objects, and no natural way of selecting one of them, then it is generally a bad idea to identify them (although sometimes this is done implicitly, as when we talk about the field with $q$ elements). This applies in particular to a vector space and its dual. Generally speaking, if you have a unique isomorphism, or even a "natural isomorphism" (this can be made precise using category theory if desired), you are safe identifying them; however, this might conceivably produce problems if they are distinct subsets of some larger set you are interested in.
To further complicate the matter, in many, or even most, cases, we don't actually care about the identity of the set in question, only its (external or internal) structure. Thus, for instance, any questions about the real numbers that cannot be answered using the fact that they form a complete ordered field (e.g., is $1 \in 2$?) are not really questions about the real numbers at all. This produces interesting issues in philosophy of mathematics; see this website for a discussion. Things get even more interesting when you start defining objects (e.g., tensor products, limits, colimits) using universal properties; in this case, the objects are only defined up to natural isomorphism, so there is no single construction that "is" the desired object.
There are also times when either of two identifications may be appropriate, but not both at once. For instance, $\mathbb{Z}$ is naturally isomorphic to a fixed subset of the padic numbers $\mathbb{Z}_p$, and to a fixed subset of the real numbers $\mathbb{R}$, but no one would ever write $\mathbb{Z} \subset \mathbb{Z}_p \cap \mathbb{R}$ (except to discuss when it is not appropriate to make identifications).
In short, it's a subtle question, and the answer is based more on experience than on any codified set of principles.
One paper I looked at needed to check every subgroup of $S_n$ for small $n$. There's a lot of symmetry there, which the author was trying to exploit. Although $S_4$ contains three subgroups isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$, the author only checked one of them; without loss of generality (``by symmetry'') that was sufficient. Problem is, one of those subgroups is normal and the other two are not, and that could have made a difference.
The punchline is that the additional case checked out finethe logical error didn't snowball.

$\begingroup$ So I guess the author of this article did not explicitely use the relation of the subgroups to $A_n$ ? $\endgroup$– ogerardMay 12, 2010 at 8:21

1$\begingroup$ Sorry for the anal heckling ten years after the fact, but $S_4$ actually has got three nonnormal subgroups isomorphic to $C_2 \times C_2$ (one for each partition of $\{1,2,3,4\}$ in two pairs). $\endgroup$ Mar 26, 2020 at 18:03
Expanding on Andy Putman's comment, beginners in linear algebra often think of all finitedimensional vector spaces (over $\mathbb{R}$, say) as the same, in particular as $\mathbb{R}^n$ with the standard basis. This can lead to mistakes in any kind of computation, say in multivariable calculus, where one has to compute simultaneously with a row vector and a column vector (e.g. an element of a vector space together with an element of its dual). The problem is that matrices act differently on row and column vectors, hence if you end up changing basis you need to deal with row and column vectors in opposite ways.

17$\begingroup$ For some unfathomable reason, mathematicians just can't spell my name correctly <grin>. $\endgroup$ May 12, 2010 at 4:46

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The answers are great so far, especially Emerton's. I would like just to add another example which I have encountered. It is the gluing construction of the structure sheaf of an affine scheme. I think the one given by "locally constant elements in the stalks" is better, but anway:
If $A$ is a ring and $X=Spec(A)$ its spectrum, then it can be shown that the assignment $\mathcal{O}_X(D(f)) := A_f$ is a sheaf on the basic open subsets. Now every sheaf on a basis of a topological space (which is closed under intersections) extends uniquely to a sheaf on the whole topological space. But wait, if $D(f)=D(g)$, it is not necessarily true that $A_f = A_g$! But we can fix that by observing that there is a unique $A$algebra homomorphism $A_f \to A_g$ if $D(g) \subseteq D(f)$, and that this is actually an isomorphism if $D(f) = D(g)$. Thus in practice it really does not matter which $g$ with $D(f)=D(g)$ is chosen.
There are some ways of formalizing this. For example you could define $Q$ to be a big ring which contains all the $A_f$ as subrings, but the identifications become identities. Define a partial order on $A$ by $f \leq g \Leftrightarrow D(g) = D(f)$. Then $A_f$ is a direct system on this partial order and we can take $Q$ as its colimit.
In Group Theory there is a notion stronger than isomorphism, i.e. similarity. For two permutation groups to be "similar" there must exist both an isomorphism between the group and a compatible (commuting in the categorical sense) isomorphism between the sets being permuted. The couple of these two isomorphisms is called a similarity. Two permutation groups can be isomorphic but not similar.
Similarity is useful to compare group constructions such as the wreath product, group presentations and classifications of subgroups of the symmetric groups.
In these situations, using only isomorphism would be a mistake and would lose information.

2$\begingroup$ The first observation is the statement that one has also to consider the corresponding element in the Burnside ring encoding distinct (formal) permutation representations of finite groups, see en.wikipedia.org/wiki/Burnside_ring. It generalizes to the fact that two linear groups can be abstractly isomorphic but can correspond to nonisomorphic representations of the underlying abstract group. $\endgroup$ May 12, 2010 at 7:01

$\begingroup$ @Roland Bacher: thanks to mention the Burnside ring. As you do, I believe this gives the proper context (and marks tables a good pedagogical tool) for similarity and isomorphism of permutation groups. $\endgroup$– ogerardMay 12, 2010 at 8:35
Spivak, Differential Geometry, Vol. 1, Chapter 3, page 16.
Background: $M$ is a manifold, $i$ is an embedding pf $M$ into a Euclidean space, and $T(M,i)$ is the tangent bundle on $M$, constructed assuming the existence of the immersion $i$, which is the way Spivak does carries out the construction at first. Then he notes that for a different choice $j$ of the immersion, we have a bundle equivalence between $T(M,i)$ and $T(M,j)$.
Then goes on(quoted verbatim):
... In other words, the dependence of $T(M,i)$ on $i$ is almost illusory; we could abbreviate $T(M,i)$ to $TM$, if we agreed that $TM$ really denotes an equivalence class of bundles, rather than one bundle. That is the sort of thing an algebraist might do, and it is undoubtedly ugly.
And then he constructs the tangent bundle in the way independent of the immersion, ie the more modern approach, and expounds its merits.
Consider $V \subset H$ both Hilbert spaces in the setting of evolution equations. It is useful to work with a Gelfand triple setting $V \subset H \equiv H^* \subset V^*$ where $H$ is identified with its dual. But then identifying $V$ with its dual is a mistake and leads to confusion.

$\begingroup$ Are you saying that if $V \subset H$, then there is a natural inclusion $H^* \subset V^*$? It seems to me that if $V \subset H$, there should be a natural surjection $H^* \to V^*$ given by restriction, $f \mapsto f_V$ for linear functionals $f \in H^*$. It is surjective by HahnBanach... no? $\endgroup$ Sep 18, 2017 at 2:48
Once I've been attending a functional analysis seminar where I've heard what seemed to be a classical pun in the field: Since all Hilbert spaces are isomorphic, all probability spaces of centered squared integrable random variables are the same, and thus probabilists should focus on, say, the uniform law on $[0,1]$ only.